Derivatives of Exponential & Logarithmic Functions - Online Tutor, Practice Problems & Exam Prep

Derivatives of General Exponential Functions

At this point in the course, we've become familiar with the process of finding a derivative, and we've learned a bunch of different rules for finding the derivatives of various functions. In this video, we're going to answer the question, "Well, how do I find the derivative of an exponential function?" That's \( b^x \), where \( b \) is just some number. It turns out that there's a rule for finding the derivative of an exponential function, just like any rule that we've seen before. Now, I'm going to give you a bit of intuition as to why this rule is what it is.

And then we'll work through some examples to actually apply this new rule. So let's go ahead and get started. We're going to take things all the way back to where we got started with derivatives at limits. So using the limit definition of the derivative, I can set up my derivative of \( b^x \) as d(bx+h)/dx as the limit as \( h \) approaches 0, bxh-bx/h.

Using my properties of exponents here, I can expand this term out, giving me the limit as \( h \) approaches 0 of \( b^x \times b^h \) using properties of exponents there, then keeping everything else the same, minus \( b^x \) over \( h \).

Looking at everything going on in this fraction here, I can see that in my numerator, these two terms both have a \( b^x \) in common. That means that I can go ahead and factor that \( b^x \) out and pull it all the way out in front of my limit. This then gives me \( b^x \times \) the limit as \( h \) approaches 0 of \( b^h - 1 \), having just factored that \( b^x \) out of both of those terms. And this is still all over \( h \). Now this limit may not look like much, but it turns out that this is actually the limit definition of the natural log.

So that then means that the derivative of \( b^x \) is \( b^x \). So pulling that term down here times the natural log of \( b \). That's exactly what this limit is. And this then gives us our rule for finding the derivative of any exponential function at \( b \) to the power of \( x \). It's \( b^x \times \ln(b) \).

One additional thing to note here is that the base \( b \) should be greater than 0, so it should be positive just because it's the base of an exponential function. And it also should not be equal to 1 because that would just make this whole derivative equal to 0. It's important to note that these restrictions are not an additional thing that you have to memorize and it really just follows from our knowledge of functions. So we want to focus here on just applying this new rule to finding the derivative of some actual functions. So let's work through some examples together here.

Looking at this first example, we want to find the derivative of \( f(x) = 6^x \). So using our new rule here to find our derivative here, \( f'(x) \), because my base is 6, my power is \( x \). That then makes my derivative f′x=6x×ln6. And this is my full derivative, \( 6^x \times \ln(6) \). Now we know that exponential functions won't always be quite this simple.

So let's look at a slightly more complicated function over here with example b. Now most of you have already learned the chain rule, but if you have not yet learned the chain rule for finding derivatives, you should focus just on applying this rule to your basic exponential functions and stop here. For the rest of you, let's get into finding the derivative of this function, \( g(x) \). This is \( 3^{x^2 + 4x} \). So we can see that this power here is not just \( x \).

It's actually a whole function of \( x \). It's \( x^2 + 4x \). So that means that we need to apply the chain rule here. Now we know that we start from the outside and work our way in. And here, we have this function inside of our exponential function.

So in finding this derivative here, \( g'(x) \), we start by applying our rule for finding the derivative of an exponential function. So this is going to be 3x2+4x×ln3, then, of course, times the natural log of our base, which in this case is 3. Now from here, we're not done yet because applying the chain rule means that we then need to multiply by the derivative of that inside function. So multiplying here by the derivative of that \( x^2 + 4x \), we know that using the power rule, this is 2x+4. And now this is my full derivative, \( g'(x) \).

Having applied the chain rule along with our rule that we just learned for finding the derivative of exponential functions. Now we're going to continue getting practice with this new rule alongside all of the rules for finding derivatives that we've already learned coming up next. I'll see you in the next one.

Find the derivative of the given function.

$f(x)=2^x-5^x$

• A. $x^2\ln2-x^5\ln⁡5$
• B. $2^{x}\ln2-5^{x}\ln5$
• C. $2^{x}\ln x-5^{x}\ln⁡x$
• D. $2\ln⁡x-5\ln x$

Find the derivative of the given function.

$g(x)=x^4+4x+4^x$

• A. $12x$
• B. $4x^3+8$
• C. $4x^3+4x^{x-1}$
• D. $4x^3+4+4^{x}\ln⁡4$

Find the derivative of the given function.

$y=(4x-3x^2+9)\cdot2^{5x}$

• A. $2^{5x}[5\ln 2(4x-3x^2+9)+(4-6x)]$
• B. $2^{5x}\cdot(-15x^2-14x+49)$
• C. $2^{5x}[(4-6x)+\ln⁡(2)(4x-3x^2+9)]$
• D. $5\ln⁡(2)(4-6x)\cdot2^{5x}$

Find the derivative of the given function.

$h\left(x\right)=4\left(\sqrt{x}+3x\right)^{\frac54}$

• A. $\frac54\left(\sqrt{x}+3x\right)\left(\frac{1}{2\sqrt{x}}+3\right)\cdot4\left(\sqrt{x}+3x\right)^{\frac54}$
• B. $\frac{5\ln\left(4\right)}{4}\left(\sqrt{x}+3x\right)\cdot4\left(\sqrt{x}+3x\right)^{\frac54}$
• C. $\frac54\cdot4\left(\sqrt{x}+3x\right)^{\frac54}\cdot\ln4\cdot\left(\sqrt{x}+3x\right)^{\frac14}\cdot\left(\frac{1}{2\sqrt{x}}+3\right)$
• D. $\left(\sqrt{x}+3x\right)^{\frac54}\cdot4\left(\sqrt{x}+3x\right)^{\frac14}$

Derivatives of General Exponential Functions Example 1

Hey, everyone. In this problem, we're asked to find the derivative of the function \( f(\theta) = \sin(\theta \times 3^{\theta^2 + 4\theta}) \), where \( \theta \times 3^{\theta^2 + 4\theta} \) is all inside the argument of my sine function here. Now it's clear that we're going to have to use multiple of our derivative rules in order to actually find this derivative. So let's get into things here.

We want to find our derivative here, \( f'(\theta) \). Remember, no matter what variable we're dealing with here, theta, x, y, whatever, we're going to find our derivative using the same exact rules. Now looking at this, my outermost function is the sine. We know that the derivative of sine is cosine, so that's where I'm going to start. So this is the \( \cos(\theta \times 3^{\theta^2 + 4\theta}) \).

But we can't stop here, right, because there's a lot going on inside of that argument. We need to apply the chain rule. Now inside the argument of my sine function, I have 2 functions being multiplied together. So while I'm applying the chain rule, I also am going to be applying the product rule. So I'm going to be multiplying by the derivative of everything that's inside of that argument here.

Now remember using the product rule, we can think of our memory tool here, "a left d right plus a right d left". So I want to take that left-hand function, which in this case is just theta. So I'm taking theta, I'm multiplying it by the derivative of \( 3^{\theta^2 + 4\theta} \). Now this is an exponential function, so I know that my derivative here is going to be that exponential function itself.

So \( 3^{\theta^2 + 4\theta} \times \ln(3) \). But because there's a lot going on in that exponent, that means I need to apply the chain rule just for this function specifically. So I further need to go ahead and multiply this by the derivative of \( \theta^2 + 4\theta \). Using the power rule, that's going to give me \( 2\theta + 4 \). Now I want to make sure that I'm keeping parentheses, keeping everything organized here.

So this is my full derivative of that \( 3^{\theta^2 + 4\theta} \). So we're still in the midst of our product rule here. So that's left d right. We're adding this together with a right d left. So keeping that right-hand function as is, that's \( 3^{\theta^2 + 4\theta} \) then times the derivative of theta, which is thankfully just 1, so multiplying that by 1.

Now we can do a little bit of cleanup here because all of these two terms have \( 3^{\theta^2 + 4\theta} \) in common. So I can go ahead and factor that out. I can also take this theta and distribute it in these parentheses here just so that I can make that a more complete polynomial with less going on outside. So I'm going to keep that cosine the same here. That's \( \cos(\theta \times 3^{\theta^2 + 4\theta}) \).

And that's the end of my argument, so I'm going to close that off with parentheses. Then this is getting multiplied here. I'm going to go ahead and factor that \( 3^{\theta^2 + 4\theta} \) out. And this is then multiplying the natural log of 3 times \( 2\theta^2 + 4\theta + 1 \) because we have factored that \( 3^{\theta^2 + 4\theta} \) out. So I can close all of my parentheses there, and this is my full derivative having applied a bunch of our rules for derivatives that we've learned up to this point.

Now I know that some of these problems can be really tricky. Make sure that you stay organized and you pay attention to every single term that's in your function. Thanks for watching, and I'll see you in the next one.

Derivatives of General Exponential Functions Example 2

Hey, everyone. In this problem, we're told that a medical lab uses a radioactive isotope for one of its tests. Now, the quantity of this radioactive isotope r remaining in a sample after t hours is given by this function: r=150×13t12. Now, we're asked to find a couple of different things here, so let's go ahead and start with part a. The first thing that we're asked to do is find the instantaneous rate of change that is dr/dt.

So, we just want to find the derivative of our function here. Now, because I have this constant 150, I can go ahead and keep that out front, and I want to multiply this by the derivative of this 13t12. Now, this is an exponential function, so I know this derivative, based on my rules for the derivatives of exponential functions, is going to be 13t12×ln 13. But because this power is t over 12, it's not just t. That means I need to take this one step further, applying the chain rule, multiplying it by the derivative of t over 12, which is in this case, 112.

Now, we can go ahead and do a little bit of simplification here. I'm going to put all of my constants together up front. So this is 150×ln 1312×13t12. So, all of this out front is just a constant.

Remember, when we take the natural log of a value, that just gives us a value. So, all of this is just a constant. And then we're multiplying this by 13t12. So, this is our rate of change of our radioactive isotope. Remember, drdt.

This indicates a time or rate of change. This is based on t hours. So now, moving on to part b, we want to compute the instantaneous rate of change at a couple of different time intervals: at after 12 hours, after one day, and after 2 days. Now in our problem, because it's indicated that t should be in hours, that means that even though we're given days here, we need to convert these into hours in order to actually plug them in. So, one day is, of course, 24 hours, and then 2 days is just that times 2, so 48 hours.

So, starting with our 12 hours, let's go ahead and plug this into our derivative. Because we're looking at the rate of change, that means we need to plug this into our derivative here. Now, I'm going to switch to prime notation here, r prime. R prime is the same thing as drdt. But since I'm plugging in these values, this is r prime of 12.

Plugging in that first value of 12 into my derivative here, this is 150×ln 1312×131212. So, all I've done here is plug 12 in for t in my derivative. Now from here, we just want to plug this all into a calculator.

And 12 over 12, we can note that this is just 1. So, when we plug this all into our calculator, if we round to 2 decimal places, this works out to be negative 4.58. Now moving on here, we want to do this for 24 hours, 48 hours. So, we want to do the exact same calculation but just plugging in 24 and 48 for t. So r prime of 24, oh, the exact same stuff here, 150×ln 1312.

And then this is 132412. So, we're plugging in 24 for t. Now we plug this into our calculator. This will again give us a negative value. It is negative 1.53.

Then finally, plugging in 48, this is r prime of 48. This is 150×ln 1312×134812. Now 48 over 12 is just 4, so you can think about that when you plug this into your calculator. Once you do plug this in, you should end up getting negative 0.17. So, we have all of these values.

These are instantaneous rates of change at each of these specific time intervals. Now here in part c, we're asked to interpret our results, which can often be kind of the hardest part of these problems, to actually think about what's going on. You may have all the math down but not know exactly what it means. Now, looking at these values, because we are looking here at an instantaneous rate of change for the quantity of a radioactive isotope after t hours, looking at these values, they're all negative. That means that the quantity of our radioactive isotope is decreasing over time.

And furthermore, if we, since we look at these values and they're getting smaller, that means it is decreasing the whole time. That's what this negative tells us, but at a slower and slower rate as time passes. So, that's exactly what these values mean. Now, if you have any questions here, feel free to let us know, and let's keep practicing.

Derivative of the Natural Exponential Function (e^x)

In the previous video, we learned a rule for finding the derivative of our general exponential function, \( b^x \). Now a common exponential function that we'll encounter is \( e^x \). So how can we find that derivative? Well, because \( e^x \) is just a special case of our general exponential function where our base is just \( e \), we can just use the derivative rule that we know for general exponential functions to then find the derivative of \( e^x \). And that's exactly what we're going to do here.

We'll also work through some additional examples together. So, let's go ahead and jump right into things. In order to find the derivative of \( e^x \), I look to my rule for general exponential functions, and I just want to replace my \( b \)'s here with \( e \)'s because that's what my base is. So, this then becomes \( e^x \), just plugging that \( e \) in, and then times the natural log of \( e \) since that's what my base is here. But the natural log of \( e \) is just 1.

So I'm just left here with \( e^x \). This means that the derivative of \( e^x \) is just \( e^x \). So, what does that look like in action? Well, coming over here, we want to find the derivative of 4 times \( e^x \). Now, using our constant multiple rule, I know that I can just pull that 4 out front, and this is \( 4 \times e^x \), which is the exact same function that we started with. Now we're going to encounter some more complicated functions that involve \( e^x \) that we'll need to find the derivative of. If you've not yet learned the chain rule and the product rule, you should stop here and just focus on these more basic functions that have \( e^x \). But for those of you that have already learned your product rule and your chain rule, let's work through some additional examples here.

The first thing we want to do is find the derivative of this function \( f(x) \). This is \( 3 \times e^{2x+4} \). So to find this derivative here, \( f'(x) \), since it starts out with this constant, I know that I can just pull that constant out front and multiply it by the derivative of \( e^{2x+4} \). Now here, we can see that that exponent is not just \( x \). It is \( 2x+4 \).

It is a function of \( x \). So that means that we need to apply the chain rule to find this derivative. So when we start with that outside exponential function, \( e^x \), we know that the derivative of \( e^x \) is \( e^x \). Since here, my exponent is not just \( x \), that means that this derivative is \( e^{2x+4} \). Then applying the chain rule, we multiply this by the derivative of that inside function.

So the derivative of \( 2x+4 \) is just 2, and that's my full derivative. Now I can write this in a bit more simple way if I go ahead and multiply those two constants together. \( 3 \times 2 \) gives me 6, so this is \( 6e^{2x+4} \) for my final derivative here, \( f'(x) \). Let's work through one additional example. Here, we want to find the derivative of \( g(x) \), and this is \( x \times e^{5x} \).

So here, I have a function \( x \) multiplying my exponential function \( e^{5x} \). That means that we need to use the product rule. Remember, our memory tool for the product rule tells us a left \( d \) right plus a right \( d \) left. So to find \( g'(x) \) here, that first term, left \( d \) right, I take my left-hand function, which is \( x \), and multiply it by the derivative of that right-hand function, \( e^{5x} \). Now again here, this power is not just \( x \).

It is \( 5x \), so we are going to use the chain rule here as well. So based on what we just saw in example a, my derivative here is gonna be \( e^{5x} \). Then using the chain rule, multiplying by the derivative of \( 5x \). The derivative of \( 5x \) is 5, and that's my full derivative of that \( e^{5x} \). Then continuing on with our product rule here, we add this together with right \( d \) left. So that's my right-hand function, \( e^{5x} \) times the derivative of \( x \), which in this case is just 1.

Now we can do some simplification here because each of these terms has \( e^{5x} \) in them. I can go ahead and just factor that out. That becomes \( e^{5x} \times (5x + 1) \).

And this is my full derivative here, \( g'(x) \), having used a lot of different rules altogether. Now we're going to continue getting practice with our new rules for exponential functions alongside all those other derivative rules that we've already learned. So let's continue on. I'll see you in the next one.

Find the derivative of the given function.

$f(x)=-3e^x+5x-2$

• A. $e^x+5$
• B. $-3e^x+5$
• C. $-3e^{x}\ln⁡3+5$
• D. $-3e^x+5x-2$

Find the derivative of the given function.

$g(x)=7e^x+2x^3$

• A. $7e^x+6x^2$
• B. $7e^x+6x$
• C. $7+6x^2$
• D. $7e^x+2x^3$

Derivative of the Natural Exponential Function (e^x) Example 3

Hey, everyone. In this problem, we're asked to find the derivative of the function \( y = e^{\sqrt{x^3 - 2x}} \). Now here, we're going to have to use the chain rule. So if you have not yet learned the chain rule for taking derivatives, go ahead and circle back to this problem after you've gone through that. But for the rest of you, let's get started with finding this derivative.

We want to find \( \frac{dy}{dx} \) here, our derivative of this function \( y \). And looking at this function, I have \( e \) raised to the power of the square root of \( x^3 - 2x \). Now because I have a function inside of a function here, I'm going to have to use the chain rule and continue multiplying by the derivative of my inside function. So starting with that outermost function, this is \( e \) raised to the power of some stuff, right?

So we're treating that stuff like it's just a variable from the beginning, and then we'll continue on working our way inside. We know that the derivative of \( e^x \) is just \( e^x \). Now since I don't just have \( x \) in that power here, this is \( e \) to the power of \( x^3 - 2x \), so exactly what is in that power here. Now is where we apply our chain rule. So we're going to multiply this by the derivative of that power.

Now we can also think of this as being \( (x^3 - 2x)^{\frac{1}{2}} \). That may help us visualize this a little bit better. So now we're just multiplying this by the derivative of \( x^3 - 2x \) to the power of \( \frac{1}{2} \). So the first thing I want to do is apply the power rule. So I pull that \( \frac{1}{2} \) out front, multiplying by it.

So \( \frac{1}{2} \times (x^3 - 2x) \), keeping that inside function the same, and then decreasing that power by 1. \( \frac{1}{2} - 1 \) gives me \( -\frac{1}{2} \). Then I'm technically going to apply the chain rule a second time, multiplying by the derivative of my innermost function, that \( x^3 - 2x \). Now the derivative of \( x^3 - 2x \) is \( 3x^2 - 2 \), applying the power rule there. So this is my full derivative.

Now we can do a little bit of cleanup here because on the top of my fraction, I have this \( e \) to the power of the square root of \( x^3 - 2x \) multiplied by \( 3x^2 - 2 \). This is in my numerator, then in my denominator. Since this is a negative power, it goes on the bottom. And I also have this one half, so that puts a 2 on the bottom there. Now \( -\frac{1}{2} \) is really the square root on the bottom here.

So I can rewrite this as \( \sqrt{x^3 - 2x} \). So this is my final derivative here, \( \frac{dy}{dx} \), having applied the chain rule here multiple times. Let me know if you have any questions, and let's continue getting practice here.

Find the derivative of the given function.

$y=x^2e^{3x^2+5x}$

• A. $2xe^{3x^2+5x-1}$
• B. $2xe^{3x^2+5x}+x^2e^{3x^2+5x}$
• C. $2xe^{3x^2+5x-1}(6x+5)$
• D. $x{e}^{3x^2+5}(6x^2+5x+2)$

Find the derivative of the given function.

$f(t)=\frac{e^{3t}}{t-2e^{-t}}$

• A. $\frac{e^{3t}(3t-8e^{-t}-1)}{(t-2e^{-t})^2}$
• B. $\frac{3e^{3t}}{1+2e^{-t}}$
• C. $\frac{(t-1)e^{3t}}{(t-2e^{-t})^2}$
• D. $\frac{3t^2e^{3t-1}-8te^{2t-1}-e^{3t}}{(t-2e^{-t})^2}$

Find the derivative of the given function.

$y=\tan \left(e^{-x^3}\right)$

• A. $\sec^2\left(e^{-x^3}\right)$
• B. $e^{-x^3}\cdot\sec^2\left(e^{-x^3}\right)$
• C. $-3x^2\cdot e^{-x^3}\cdot\sec^2\left(e^{-x^3}\right)$
• D. $-x^3\cdot e^{-x^3-1}\cdot\sec^2\left(e^{-x^3}\right)$

Derivative of the Natural Exponential Function (e^x) Example 4

Hey, everyone. In this problem, we want to find an equation to the tangent line of the function y=3×ex−x at x=0. So how do we find the equation of a tangent line? Well, we need to find the slope of that tangent line, which is, of course, going to be our derivative. So starting there, dy/dx, our derivative of 3×ex−x.

Since I have two functions here being subtracted, I can just subtract their individual derivatives. So the derivative of 3 times ex is just 3 times ex. It's the same exact thing. And the derivative of negative x is just negative one. So my derivative here, 3×ex−1.

But since here I'm working specifically at x=0, my slope at this exact point means that I need to go ahead and plug a 0 in for x. So plugging 0 in for x here, that's 3 times e0−1. Anything to the 0 is just one. So this is really just 3 minus 1, which is 2. So this here is my slope.

But we don't just want to find the slope. We want to find the entire equation of the tangent line. So how can we do that? Well, remember, our slope-intercept form tells us y−y1=m×(x−x1). Now we already have our slope here.

We know that it's just 2. We also know that x1=0. So the only thing that we have left to find here is a y1. So that means we need to plug 0 into our original function. So if we go ahead and do that here in order to find y1, this is going to be 3 times e0−0, just plugging 0 in for x.

So this just gives us 3 because subtracting 0 doesn't do anything. And e0 is just 1. So now we can plug in a y1, x1, and m. So this is a y−y1=2×(x−x1).

Now we can rewrite this. If we move this 3 over to the other side and we go ahead and distribute this 2, this then becomes y=2x+3. So this is the equation of the tangent line to the function y=3×ex−x at x=0. Let us know if you have any questions here and let's keep going.

Derivatives of General Logarithmic Functions

Hey, everyone. We just learned a rule for finding the derivative of our general exponential function, \( b^x \). Here, we're going to take a look at finding the derivative of our general logarithmic function. That's \( \log_b(x) \). Now there is, of course, a rule that we can use in order to find this derivative.

Now here, I'm going to walk you through sort of where this rule comes from using what we already know about exponential functions to give you a little bit of intuition as to why this rule is what it is. Once we actually know this rule, we can just apply it to some examples, and we'll work through those together. So let's go ahead and jump into things. Now looking at this equation, \( y = \log_b(x) \), my ultimate goal here is to determine my derivative, \( \frac{dy}{dx} \). We want to find what that is.

Now this equation is currently written in its logarithmic form, but I can rewrite this in its equivalent exponential form using properties of logs and exponents, taking this base and raising it to the power of \( y \). This is then equal to \( x \). So I can rewrite this as \( x = b^y \). Now since this is an exponential function, we can then use what we already know about finding the derivative of exponential functions. So we can differentiate both sides of this equation in order to ultimately find the derivative of our logarithmic function that we're looking for.

That's \( \frac{dy}{dx} \). So having rewritten \( y = \log_b(x) \) as \( x = b^y \), we now want to take the derivative on both sides here. Now if you're familiar with the process of implicit differentiation, that's exactly what we're doing here. But if you aren't familiar with that process, all we're doing is just taking the derivative on either side using all the rules that we already know, so don't worry about it. Now looking at this, the derivative of \( x \) with respect to \( x \), we know that this is just equal to 1.

Then coming over to our right side, the derivative of \( b^y \), I know that I can use my rule for finding the derivative of an exponential function here. But since this power is not \( x \), it is \( y \), that means I need to apply the chain rule when I'm taking this derivative. So I have \( b^y \times \ln(b) \) using my rule here. And then I want to multiply this by the derivative of that inside function \( y \) since I know that that is a function of \( x \). This is \( \frac{dy}{dx} \).

Now this is ultimately the derivative that we want to find. We want to find the derivative of our logarithmic function. That's \( \frac{dy}{dx} \). So we want to solve for this \( \frac{dy}{dx} \), which we can do by isolating it, dividing this over, this \( b^y \times \ln(b) \), in order to get it to cancel. Then we're just left with that \( \frac{dy}{dx} \).

Now swapping the sides of my equation here, this gives me that \( \frac{dy}{dx} = \frac{1}{b^y \times \ln(b)} \). And this is the exact derivative that we were looking for. Now, typically, when we're working with derivatives with respect to \( x \), we don't usually want that derivative in terms of \( y \). But we know exactly what \( y \) is. It's \( \log_b(x) \).

So we can just plug that right in here. So this is \( b^{\log_b(x)} \times \ln(b) \), just having replaced that \( y \) with this \( \log_b(x) \). But using properties of logs and exponents, if I take \( b \) and raise it to the power of \( \log_b \), that just cancels out. So all I'm left with here is just \( x \).

So then \( \frac{dy}{dx} \), my derivative of \( \log_b(x) \), is equal to \( \frac{1}{x \times \ln(b)} \). That's what's left over in my denominator here. And this is my full rule for finding the derivative of a general logarithmic function. The derivative of \( \log_b(x) \) is \( \frac{1}{x \times \ln(b)} \). Now these restrictions on \( b \) are the same exact ones that we saw when dealing with our exponential function.

And here, \( x \) must also be greater than 0 because we know we can't take the log of a negative number, and it does end up in the denominator of my fraction. So, hopefully, this gave you a little bit of context as to why this rule is what it is. But now that we have this rule, we can just apply it to some examples and actually put it to use. So let's work through our first example here and find the derivative of this function, \( g(x) = \log_8(x) \). So applying my rule here in order to find \( g'(x) \), my derivative, this is then going to be equal to \( \frac{1}{x \times \ln(8)} \).

And this is my full derivative. \( g'(x) = \frac{1}{x \times \ln(8)} \). Now we're going to have to deal with finding the derivative of some more complicated functions involving these logarithmic functions. If at this point you have not yet learned the chain rule, you should stop here and just focus on applying this rule to some basic logs. But for the rest of you, let's continue on in finding the derivative of this next function, \( f(x) \).

This is \( \log_5(x^2) \). So here, we can see that we're not just taking \( \log_5(x) \). We're taking \( \log_5(x^2) \). Because of that, we need to use the chain rule because we have a function inside of our logarithmic function. So in order to find this derivative, \( f'(x) \), we're going to use the rule that we just learned but along with the chain rule.

So when starting with this outside function, the derivative of a \( \log_5(x^2) \), when we take this derivative, we cannot just put an \( x \) there. We have to put the full function that we started with. So this is then \( \frac{1}{x^2} \), not just \( x \), times the natural log of our base, which is just 5. Then from here, applying the chain rule, we multiply this by the derivative of that inside function. Our inside function is, in this case, \( x^2 \).

So the derivative of \( x^2 \) is just 2x. So we multiply this by 2x. That allows these \( x \)'s to cancel, and I am left here with \( \frac{2}{x \times \ln(5)} \) for my final derivative here, \( f'(x) \), having applied both the chain rule and our new rule for finding the derivative of logarithmic functions. Now we're going to continue getting practice putting all of our rules for derivatives all together. I'll see you in the next one.

Find the derivative of the given function.

$f(x)=8\log_2⁡x$

• A. $\frac{8}{2\ln x}$
• B. $\frac{8}{x\ln2}$
• C. $\frac{8}{x}$
• D. $\frac{1}{x\ln8}$

Find the derivative of the given function.

$g(t)=2t+\log_5⁡t$

• A. $\frac{2}{t\ln5}$
• B. $\frac{1}{t\ln5}$
• C. $2+5\ln t$
• D. $2+\frac{1}{t\ln5}$

Derivatives of General Logarithmic Functions Example 5

Here, we're asked to find the derivative of the function \( y \) equals \( \log_4 \left( \sqrt{\frac{x}{4x+1}} \right) \). Now from just taking a look at this function, we could see that there's quite a bit going on here. In order to take this derivative, we're going to have to use a bunch of different rules altogether. So we're going to use our rule for finding the derivative of a general logarithmic function, but also our chain rule since we're not just taking the log base 4 of \( x \), we're taking the log base 4 of \( \sqrt{\frac{x}{4x+1}} \). And in using the chain rule, we're also going to have to apply the quotient rule.

Now you likely have already learned both the chain rule and the quotient rule, but if you haven't, go ahead and learn those rules before circling back to this problem. So let's go ahead and get into things here. We want to find our derivative, \( \frac{dy}{dx} \). Now in order to find this derivative here, we're starting with our outermost function. This is this log base 4.

Now we know that for a general logarithmic function, our derivative is going to be \( \frac{1}{x} \times \ln(\text{our base}) \). Since we're not taking the derivative of just \( x \) here, that makes this derivative \( \frac{1}{\sqrt{\frac{x}{4x+1}}} \times \ln(4) \). Now this is our outermost function. Using our chain rule, we need to work our way inside, continuing to multiply by each successive derivative. So now looking at this inside function, that's the \( \sqrt{\frac{x}{4x+1}} \).

We're multiplying by the derivative of this function. Now it can be helpful here to think of this function as \( \left(\frac{x}{4x+1}\right)^{0.5} \), so it makes it a bit more clear how we're applying the power rule here. Now in applying the power rule, we're pulling that \( 0.5 \) out to the front and then decreasing it by 1. So this is \( 0.5 \times \left(\frac{x}{4x+1}\right)^{-0.5} \). So that means that this is on the bottom of a fraction, and it is a square root.

Now we're not quite done yet because we have started with our outermost function. We found the derivative of that log base four term. Now we're working our way inside. We took the derivative of this \( \sqrt{\frac{x}{4x+1}} \). But because that inside function is \( \frac{x}{4x+1} \), we need to do one more step here and multiply by the derivative of that fraction.

Now since I have one function dividing another function, I need to use the quotient rule within this chain rule that I'm doing. So remember our chain rule memory tool, low \( d \) high minus high \( d \) low over the square of what's below. That's what's going to give us this derivative of \( \frac{x}{4x+1} \). So low \( d \) high, so my low function, \( 4x+1 \), times the derivative of that function on top. The derivative of \( x \) is just 1 minus high \( d \) low, so that's \( x \), times the derivative of \( 4x+1 \), that's just 4, over the square of what's below, so that's \( (4x+1)^2 \).

And make sure you square that term. Now from here, we've taken all of our derivatives, and we just have to do some simplification. So we have the derivative of our log function. We have the derivative of that square root, and we have the derivative of what's inside of that square root. So we worked our way from outside to inside using the chain rule.

Now from here, we're going to look at everything that's going on. I can see that in this numerator, I have a positive 4x and I have a negative 4x. That means those 2 are going to cancel each other out. So looking at what's actually left in my numerator here, because this is going to go on the bottom, that's going to go in my denominator as well. The only thing I actually have in my numerator is 1, so my numerator just becomes 1.

My denominator is a little bit more complicated. So I'm going to go ahead and pull my constants out front. So I have 2 and I have \(\ln(4)\). And I keep those out here, \( 2 \times \ln(4) \). Then multiplying this by this term, so that's the \( \frac{x}{4x+1} \).

Then this is also the square root of \( \frac{x}{4x+1} \). It's on the bottom again because of that negative exponent. So this is the same exact term. I've accounted for that one. And then finally, I just have to account for this \( 4x+1 \) squared.

But we can do some massive simplification here because I have these 2 square roots multiplying each other. That means the square roots are going to cancel out because, effectively, I have the square root of \( \frac{x}{(4x+1)^2} \). So those square roots will cancel out and leave me with just one of those terms. So this is \( \frac{1}{2 \times \ln(4)} \), that stays the same, times \( \frac{x}{4x+1} \) because those 2 multiplied together have just given me that inside term. Then finally, I have times \( 4x+1 \) squared.

And looking at this, I can simplify this even further because I can cancel one of these \( 4x+1 \) with one of these \( 4x+1 \). Then finally, what am I left with? I'm left with a \( \frac{1}{2 \times \ln(4)} \) and then times \( x \times (4x+1) \). Now, you can rewrite this a little bit nicer just so that it's clear that we're only taking the \(\ln(4)\). So this is really \( 2x \ln(4) \).

This is in my denominator. So this is \( \frac{1}{2x \ln(4) \times (4x + 1)} \). So this is my full derivative here, \( \frac{1}{2x \times \ln(4) \times (4x+1)} \). Now often, the messiest part of these problems is the algebra that you get to, and I know that some of these can be tricky. So make sure that you go ahead and account for every single term each time you start a new line.

Let us know if you have any questions, and I'll see you in the next one.

Find the derivative of the given function.

$f(x)=(x^3+2x)\cdot\log_5⁡x$

• A. $\frac{(3x^2+2)}{x\ln5}$
• B. $(3x^2+2)\cdot\log_5⁡x+x^2+2$
• C. $\left(3x^2+2+\frac{x^2}{\ln5}+\frac{2}{\ln5}\right)\cdot\log_5x$
• D. $\frac{x^3+2x}{x\ln5}+\log_5x\cdot\left(3x^2+2\right)$

Find the derivative of the given function.

$g\left(t\right)=\log_5\left(7^{t^2+4}\right)$

• A. $\frac{2t\ln7}{\ln5}$
• B. $2t\ln\frac75$
• C. $\frac{2t\ln7}{t\ln5}$
• D. $\frac{7^{t^{2+4}}}{t\ln5}$

Find the derivative of the given function.

$h\left(x\right)=\sin\left(\log_4\left(x^2\right)\right)$

• A. $\frac{\cos\left(\log_4x^2\right)}{x^2\ln4}$
• B. $\frac{2\cos\left(\log_4x^2\right)}{x\ln4}$
• C. $\cos\left(\log_4x^2\right)$
• D. $\frac{2\cos\left(\log_4x^2\right)}{\ln4}$

Derivative of the Natural Logarithmic Function

Hey, everyone. In the previous video, we learned a rule for finding the derivative of our general logarithmic function. That's log base b of x. Now a common logarithmic function that we'll encounter is the natural log of x. So how can we find this derivative?

Well, since the natural log of x is just a special case of our general logarithmic function where our base is just e, that means that we can then use the derivative rule that we already know for general logarithmic functions in order to then find the derivative of the natural log, that is log base e of x. Now that's exactly the process that we're going to do here. Once we then know the derivative of the natural log, we can work through some additional examples together. So let's go ahead and get started here. So here, I have my derivative of the natural log of x.

Now since this is just log base e of x, I can then rewrite this. Then I can apply my rule for my general logarithmic function, just replacing that b with an e because that's what my base is here. So this becomes 1 / x × ln(e). But we know that the natural log of e is just equal to 1. So that leaves us here with just 1 / x.

And that's exactly what my derivative is. The derivative of the natural log of x is 1 / x. Now, the only restriction that we have to worry about here is that x is greater than 0 because we no longer have to worry about b. So now that we know that the derivative of the natural log is 1 / x, let's take a look at an example. Here, we want to find the derivative of 6 × ln(x).

Now since 6 is just a constant, I know that I can pull it out front of my derivative using the constant multiple rule. So this is then just 6 × ddx of ln(x). Now we just found that the derivative of the natural log of x is just 1/x. So this is then 6/x. And this is my completed derivative.

Now at this point, if you have not yet learned the chain rule or the product rule, you should stop here and just focus on applying this rule to some basic functions involving the natural log. For the rest of you, we're going to take a look at a slightly more complicated example here in finding the derivative of this function, f(x). Now this function is the natural log of x2 + 4x. So we can see here that we're no longer just taking the natural log of x but the natural log of x2 + 4x. So that's an entire function of x inside of my natural logarithmic function.

That means that I, of course, need to apply the chain rule here. Now when applying the chain rule, we work from the outside in. So to find our derivative here, f'ofx, we're going to start with that outermost function, which here is the natural log. So we know that the derivative of the natural log is just 1/x. But here, since we have a function that we're taking the natural log of, that means that when we apply the chain rule, this is not just 1/x, but one over that entire function of x.

So in this case, that is 1/x2 + 4x. Then further applying the chain rule here, we, of course, then need to multiply by the derivative of that inside function. So the derivative of x2 + 4x is 2x+4, and this is our full derivative. And we can write this a little bit more compactly with just 2x+4 over that x2 + 4x. And this gives us our final derivative here, f'ofx, having applied the chain rule along with our rule for our natural logarithmic function.

Now, let's take a look at one more example here. Here, we want to find the derivative of this function, g(x), which is x × ln(x3). So here, since I have this function x multiplying the natural log of x3, that means I need to use the product rule. Now remember that our memory tool for the product rule tells us left d right plus a right d left. So g'ofx here, I'm going to start with that left-hand function, which in this case is x, and multiply that by the derivative of my right-hand function, the natural log of x3.

Now here, again, we can see that we're not just taking the natural log of x. We're taking the natural log of x3. So we need to attack this the same way that we did in example a. So in finding the derivative of this natural log of x3, I start with that outermost function. I know that the derivative of the natural log is a 1/x.

But here, since I have x3, that makes this 1/x3. And then I need to multiply by the derivative of that inner function x3, which is 3x2. Then continuing on with our product rule here, that was left d right. We're adding this together with a right d left. So that right-hand function, the natural log of x3 as is, and then multiplying that by the derivative of my left-hand function.

The derivative of x is just 1. Now from here, we can do a bit of cancellation because this x3 on the bottom is going to cancel out with that x × x2 on the top. That just leaves me with 3 + ln(x3). So this is my full derivative at g'ofx. Now we can see that we put a lot of different derivative rules altogether in that example.

And we're going to continue getting practice with that coming up next. Let me know if you have any questions, and I'll see you in the next one.

Find the derivative of the given function.

$f(x)=2x^3-1+\ln x$

• A. $6x$
• B. $\frac{3x^2-1}{x}$
• C. $6x^2+\frac{1}{x}$
• D. $6x-1$

Find the derivative of the given function.

$g(x)=e^{x}+\ln x^5$

• A. $\frac{5e^{x}}{x}$
• B. $e^{x}+5\ln x$
• C. $\frac{e^{x}+5}{x}$
• D. $e^{x}+\frac{5}{x}$

Find the derivative of the given function.

$h\left(x\right)=\ln\left(\frac{\sqrt{x+1}}{x^2+3}\right)$

• A. $\frac{-3x^2-4x+3}{2\left(x+1\right)\left(x^2+3\right)}$
• B. $\frac{x^2+3}{\sqrt{x+1}}$
• C. $\frac{-3x^2-4x+3}{2x\sqrt{x+1}\cdot\left(x^2+3\right)^2}$
• D. $\frac{-3x^2-4x+3}{2\left(x+1\right)\left(x^2+3\right)}\cdot\ln\left(\frac{\sqrt{x+1}}{x^2+3}\right)$

Find the derivative of the given function.

$y=x^2\ln⁡(x^2)$

• A. $2x\ln x^2+\frac{2}{x}$
• B. $2x\left(\ln\left(x^2\right)+1\right)$
• C. $\frac{2}{x}$
• D. $\frac{1}{x^2}$

Find the derivative of the given function.

$g(x)=e^{x^2\ln x}$

• A. $e^{x^2\ln x}$
• B. $e^{2x+\frac{1}{x}}$
• C. $xe^{x^2\ln x}\left(2\ln x+1\right)$
• D. $2e^{x^2\ln x}$

Derivative of the Natural Logarithmic Function Example 6

If you haven't already given this problem a try on your own, go ahead and do that before checking back in with me. Alright. In this problem, we're asked to use implicit differentiation to find \( \frac{dy}{dx} \). And we're given here that the natural logarithm of \( x^2 \times y \) is equal to \( 2 \times e \) to the power of \( 3x + y \). So I'm going to start here by just rewriting my equation.

That's the natural logarithm of \( x^2 \times y \), and this is equal to \( 2e^{3x+y} \). So in order to use implicit differentiation to ultimately find \( \frac{dy}{dx} \), we need to take the derivative with respect to \( x \) of both sides here. So I'm going to start on this left-hand side. I want to take the derivative of the natural logarithm of \( x^2y \) with respect to \( x \). Now in doing this, this is going to give me \( \frac{1}{x^2y} \).

And then applying the chain rule, I then have to multiply this by the derivative of \( x^2y \). Now in order to find this derivative, we need to use the product rule because I have \( x^2 \) multiplying \( y \). So that means here that we need to apply our product rule. Remember our memory tool for the product rule, left \( d \) right plus a right \( d \) left. So I'm going to take that left-hand function \( x^2 \), multiply it by the derivative of that right-hand function \( y \), which is \( \frac{dy}{dx} \).

And I'm adding this together with that right-hand function \( y \) times the derivative of that left-hand function \( x^2 \), which is just \( 2x \). So this is my left-hand side, that whole derivative. This is then equal to the derivative of \( 2 \times e \) to the power of \( 3x+y \). That \( 2 \) is just a constant, so I can leave that out front. Then the derivative of \( e^{3x+y} \) is then \( e^{3x+y} \times \) the derivative of \( 3x+y \).

The derivative of \( 3x+y \), because we're using the chain rule here, is just going to be \( 3+\frac{dy}{dx} \). So now we have fully taken these derivatives, and we really just have to do a bunch of algebra from this point on because we want to find \( \frac{dy}{dx} \). So to ultimately get \( \frac{dy}{dx} \), since it's in 2 different terms, we need to first start by expanding this even further so that we can then simplify it down as we go on. So the first thing I want to do is actually just distribute everything into my parentheses here. So starting on this left-hand side, this gives me \( \frac{x^2}{x^2y} \times \frac{dy}{dx} + \frac{2xy}{x^2y} \).

Then this is equal to \( 6e^{3x+y} + 2e^{3x+y} \times \frac{dy}{dx} \). So now I can see that I have 2 terms that have this \( \frac{dy}{dx} \). I want to get those on the same side so that I can ultimately solve for that \( \frac{dy}{dx} \). So I want to pull this term over to this left side subtracting by it, and that means that I also want to take my terms that don't have \( \frac{dy}{dx} \) and put them on the same side as well.

So I'm going to subtract this term over to the right side. So cleaning this up a little bit as I go, I'm going to go ahead and cancel these \( x^2 \). So this gives me \( \frac{1}{y} \times \frac{dy}{dx} \). And then subtracting this term over here, this is \( -2e^{3x+y} \times \frac{dy}{dx} \). This is then equal to that \( 6e^{3x+y} - \frac{2}{x} \).

And I can do some cancellations here as well. Those \( y's \) cancel, and then one of my \( x's \) cancels as well. So this is just leaving me with \( \frac{2}{x} \) for that term. So looking at this left-hand side, I can factor out \( \frac{dy}{dx} \) since both of those terms have it. So that's \( \frac{dy}{dx} \times \left(\frac{1}{y} - 2e^{3x+y} \)

So all I've done is I've pulled that \( \frac{dy}{dx} \) out of both of those terms. Then on my right-hand side, there's nothing else that I need to do here yet. This is \( \frac{6e^{3x+y} - \frac{2}{x}} \). So now that I have this \( \frac{dy}{dx} \) factored out, in order to isolate it, all I need to do is get rid of this term, which I can do by dividing it over. So if I divide by it on both sides, so this is \( \frac{1}{y} - 2e^{3x+y} \).

That's going to cancel over here. So I'm dividing by it over here. This is \( \frac{1}{y} - 2e^{3x+y} \) again there. Then on my left-hand side, all that I'm left with finally is \( \frac{dy}{dx} \). But we definitely have a bit of algebra to do over here before this is fully simplified.

And in order to get through this algebra, since I have a fraction on top and I have this fraction on a bottom, but I want to go ahead and have my other term have a common denominator so that I actually can subtract them, and it will ultimately make it easier to simplify this further. So for this \( 6 \times e^{3x+y} \), I'm going to multiply this by \( \frac{x}{x} \) to give it that common denominator of \( x \). Then on the bottom, since this fraction, one over \( y \), has a denominator of \( y \), I'm going to multiply that other term by \( \frac{y}{y} \) to give it a common denominator as well. Now when we do this, this becomes \( \frac{6xe^{3x+y}-\frac{2}{x}} \) because I've given it that common denominator. And that's the numerator of this entire fraction.

And then on the denominator there, this then becomes \( \frac{1 - 2y \times e^{3x+y}}{y} \) because, again, I've given those a common denominator as well. Now here, since I'm dividing 1 fraction by another fraction, I am ultimately just multiplying this fraction by the reciprocal of this fraction. So what is this actually equal to? Well, we can rewrite this as \( \frac{6x \times e^{3x+y} - \frac{2}{x}} \). So that is my full numerator there.

And since I'm dividing by a fraction, I'm multiplying by the reciprocal of it. So this is then multiplied by \( \frac{y}{1 - 2y \times e^{3x+y}} \). So now I can just distribute this \( y \) into these two terms and distribute this \( x \) into these two terms to get my final answer. So here I have \( \frac{dy}{dx} \) is then equal to, distributing that \( y \) into those top two terms, \( \frac{6xy \times e^{3x+y} - \frac{2y}{x}} \) minus \( 2xy \times e^{3x+y} \). And this is my final simplified answer here.

Now I know that a lot of this algebra can be a little bit tedious to get through. Make sure that you're paying attention to every single term that is in each line as you work through these problems. I hope that this was helpful. And if you have any questions, let us know.

Derivative of the Natural Logarithmic Function Example 7

If you haven't already given this problem a try on your own, go ahead and do that before checking back in with me. Alright. Here, we're told that the time \( t \) in days that it takes the number of bacteria in a sample to reach \( b \) is given by the function below. So this is \( t(b) \), and it's equal to \( 42 \times \ln\left(\frac{650}{1500 - b}\right) \). Now we're asked here to find \( t'(1000) \) and then explain what it represents.

So before we get to that explanation piece, let's focus on the calculus at hand. We're asked to find \( t'(1000) \). So we want to find the derivative of this function and then plug 1000 in for \( b \). Let's start with the derivative piece, so \( t'(b) \). Since I have this constant out front, I'm going to leave that out front.

This is \( 42 \times \) the derivative of \( \ln\left(\frac{650}{1500 - b}\right) \). Multiplying by this derivative, the derivative of the natural log is \( \frac{1}{\text{whatever that argument is}} \), and then we need to multiply using the chain rule by the derivative of what's inside of this. Here, if we take one over this fraction, it really just flips the fraction. So this becomes \( \frac{1500 - b}{650} \). Then we multiply this by the derivative of that inside function.

You can think of this function as \( 650 \times (1500 - b)^{-1} \) if you don't want to use the quotient rule. But if you do want to use the quotient rule, that's the method that I'm going to use here. Either one should get you the correct answer. When taking this derivative using the quotient rule, remember, low \( \text{d high} - \text{high d low} \) over the square of what's below.

So my low function, that's \( 1500 - b \), times the derivative of that top function. Because it's just a constant, that's going to be 0, minus high \( \text{d low}, \) so that's 650, times the derivative of \( 1500 - b \), which is just \( -1 \), over the square of what's below. That's \( (1500 - b)^2 \). Now looking at this, there are some things that can cancel out. This \( 0 \) goes away because all of that is just 0 since it's being multiplied.

Then I can see on the top here, I have \( 1500 - b \). I have \( (1500 - b)^2 \) down here. So I can cancel one of those out. I also have a \( 650 \) on top, a \( 650 \) on the bottom, so those would go away. And I'm left on the top here with \( 42 \) times negative negative one.

That's just positive one, so I'm left on the top with just \( 42 \) and then on the bottom with just \( 1500 - b \). This is my derivative here. I found \( t'(b) \). Now we want to find \( t'(1000) \) by plugging \( 1000 \) in for \( b \). So \( t'(1000) \), this is \( \frac{42}{1500 - 1000} \).

\( 1500 - 1000 \) just gives us 500. So this is \( \frac{42}{500} \) or as a decimal, 0.084 if we go ahead and plug that into our calculator. So this is \( t'(1000) \). Now remember that the final piece of this problem is to explain what this value represents. And this can often be the hardest part of these types of problems.

You know all of the calculus, but it may be hard to interpret exactly what's going on in the problem. Now remember that we took a derivative here, and a derivative is a rate of change. Here it can feel a little bit backward because we took the derivative of time with respect to bacteria rather than with bacteria with respect to time. So we really calculated \( dtdb \). It can be helpful whenever you're interpreting results like this to think about what units we're dealing with.

So \( t \) was measured in days. \( b \) is our bacteria. Now we plugged in 1000 for \( b \). So this means that when our sample has 1000 bacteria present, it is taking 0.084 days per bacteria. This is our growth rate here.

So in other words, if we think about this instead of in days, if we convert this value into hours, we would find that, really, one bacteria every 2 hours is growing because it could make more sense to think about this as bacteria over days as opposed to the other way around. But if we look at these results as is, when our sample has 1000 bacteria present, it is growing at a rate of 0.084 days per new bacterium. I hope that this was helpful, and let us know if you have any questions here. I'll see you in the next one.

Graphical Applications of Exponential & Logarithmic Derivatives: Example 8

Here we're asked to find the critical points of the given function. The function that we're given here is \( y = \ln(x - 1) \). Now remember that in order to find critical points, we're looking for where our derivative, so in this case, \( y' \), is either equal to 0 or does not exist. So the first thing we want to do here is find the derivative. Finding our derivative here, \( y' \), we know that the derivative of the natural log is \( \frac{1}{\text{argument of the natural log}} \).

Now here, that is \( \frac{1}{x - 1} \). Using the chain rule here, I don't actually need to do anything because the derivative of that inside function is also 1. So my derivative is just \( \frac{1}{x - 1} \). We want to find where this derivative is either equal to 0 or does not exist. Now there are actually no values for which this is going to be equal to 0 because there's nowhere that our numerator is going to be equal to that.

So we don't need to even think about this being equal to 0. We instead need to focus on where it does not exist. And this is specifically going to be where our denominator is equal to 0 because we know that a denominator cannot be 0. So if I set that denominator \( x - 1 = 0 \), that then gives me, adding 1 to both sides, \( x = 1 \). So this may be a critical point.

But remember that whenever we're working with these, we need to make sure that they're inside of the domain of our original function. If I were to plug 1 into my original function, that would give me \( \ln(0) \), which is not something that I can do. So because this one is not within the domain of my original function, I just don't have any critical points. So for this function, \( \ln(x - 1) \), there are actually no critical points at all. Now feel free to let us know if you have any questions here.

Thanks for watching, and I'll see you in the next video.

Graphical Applications of Exponential & Logarithmic Derivatives: Example 9

In this problem, we're asked to find the absolute maximum and minimum values of the function over the given interval. Now the function that we're working with here is f(x)=x⋅ln(x) over the given interval from 1 to 2. With both of those endpoints included, this is a closed interval indicated by those square brackets. So based on the extreme value theorem and what I see with this function and its closed interval, I know that this function does have an absolute max and an absolute min, which I can find by starting with my steps here. Now the first thing I want to do is find my critical points.

That's where my derivative is either equal to 0 or does not exist. So the first thing I want to do is find my derivative f′(x). Now since I have one function, x, multiplying another function, the natural log of x, I want to use my product rule, which if we remember our little memory tool here, that's a left d right plus a right d left. So I have my left-hand function, that's x, times the derivative of that right-hand function. The derivative of the natural log of x is 1x.

Then I'm adding that together with my original function over here, natural log of x times the derivative of x, which is just 1. Now I can do some simplification here. These x's cancel, leaving me with 1+ln(x) because natural log of x times 1 is just the natural log of x. So here I have my derivative. Remember, we want to look for where this derivative is equal to 0 or where it does not exist.

Now the place where this derivative does not exist is where x is equal to 0 because we can't take the natural log of 0. Right? But 0 is not within that interval. So even though one of my critical points is x equals 0, it doesn't matter because it's not in that interval. So we now just need to set this equal to 0 and find our other critical points.

So if I go ahead and subtract 1 from both sides, that gives me ln(x)=−1. And if I go ahead and raise both sides with e, these e and natural log cancel out, leaving me with x=e−1, which we could also rewrite this as 1e. So this is our other critical point. We have found both of our critical points. We know that one of them is already not one we're going to use.

And from here, we want to plug our critical points and our endpoints into our original function. Remember, we need to double-check that our critical points are in that included interval. It can be hard to tell if something like one over e is within a specified interval. But if we go ahead and type one over e into our calculator, that's going to give us a value of about 0.37, which is not within that established interval. So I don't need to plug this critical point in either.

And all I need to do here is actually just plug my endpoints in because none of my critical points are within my interval. So I'm going to go ahead and plug my endpoints in. That's x equals 1 and x equals 2. So plugging those in to my original function, which is x times the natural log of x, this gives me here f(1) as a 1 times the natural log of 1, which is going to give me a value of 0. And then for 2, plugging 2 in, that's 2 times the natural log of 2.

Typing that into my calculator gives me about 1.39. So I only have 2 values here. One of them is the largest. One of them is the smallest. That's going to give me my max and my min over this closed interval.

So I can see here that this 1.39 is my largest value at x equals 2. So this represents the global max of my function over this interval. And then at x equals 1, my function value is 0. That's my smallest value. So this is my global minimum.

So we have our global minimum of 0 at x equals 1, our global max of 1.39 at x equals 2. Feel free to let us know if you have any questions, and I'll see you in the next video.

Identify the open intervals on which the function is increasing or decreasing.

$f(x)=xe^{-2x}$

• A. Increasing on $\left(-\infty,\frac12\right)$, Decreasing on $\left(\frac12,\infty\right)$
• B. Increasing on $\left(\frac12,\infty\right)$$\left(-\infty,\frac12\right)$, Decreasing on $\left(-\infty,\frac12\right)$
• C. Increasing on $\left(-\infty,\infty\right)$
• D. Decreasing on $\left(-\infty,\infty\right)$

Identify the local minimum and maximum values of the given function, if any.

$f(t)=t^2\ln t$, $t>0$

• A. Local maximum of $1$ at $x=1$, Local minimum of $-0.19$ at $x=\frac12$
• B. Local maximum of $-\frac{1}{2e}$ at $x=\frac{1}{\sqrt{e}}$ , No local minima
• C. Local minimum of $-\frac{1}{2e}$ at $x=\frac{1}{\sqrt{e}}$ , No local maxima
• D. No local extrema

For the following graph, find the open intervals for which the function is concave up or concave down. Identify any inflection points.

• A. Concave down: $\left(0,\infty\right)$; No inflection pt
• B. Concave down: $\left(0,\infty\right)$; Inflection pt: $\left(0,0\right)$
• C. Concave up: $\left(-\infty,0\right)$; Concave down: $\left(0,\infty\right)$; Inflection pt: $\left(0,0\right)$
• D. Concave down: $\left(-\infty,\infty\right)$; No inflection pt

Determine the intervals for which the function is concave up or concave down. State the inflection points.

$f(x)=4\ln(3x^2)$

• A. Concave down: $\left(-\infty,0\right)$; Concave up: $\left(0,\infty\right)$; Inflection pt: $\left(0,0\right)$
• B. Concave down: $\left(-\infty,0\right)$ & $\left(0,\infty\right)$; No inflection pt
• C. Concave up: $\left(-\infty,0\right)$; Concave down: $\left(0,\infty\right)$; Inflection pt: $\left(0,0\right)$
• D. Concave up: $\left(-\infty,0\right)$; Concave down: $\left(0,\infty\right)$; No inflection pt