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4. Derivatives of Exponential & Logarithmic Functions

Derivatives of Exponential & Logarithmic Functions

4. Derivatives of Exponential & Logarithmic Functions

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

1

concept

Derivatives of General Exponential Functions

Video duration:

4m

Video transcript

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 to the power of 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 other rule that we've seen before. Here, 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. 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. Using the limit definition of the derivative, I can set up my derivative of bx as the limit as h approaches 0, bx+h-bxh. Using my properties of exponents here, I can expand this term out, giving me the limit as h approaches 0 of bxbh, then keeping everything else the same, minus bx over h.

Now looking at everything going on in this fraction here, I can see that in my numerator, these two terms both have a bx in common. That means that I can go ahead and factor that bx out and pull it all the way out in front of my limit. This then gives me bx times the limit as h approaches 0 of bh-1h. 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 bx is bx. 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 bx. It's bx times the natural log of b.

One additional thing to note here is that the base b should be greater than 0, as it should be positive since 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. 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 of x equals 6 to the power of x. So using our new rule here to find our derivative here, f prime of x, because my base is 6, my power is x. That then makes my derivative 6x times the natural log of my base here, which again is 6. And this is my full derivative, 6x×ln(6). Now we know that exponential functions won't always be quite this simple.

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 of x. This is 3 to the power of x squared plus 4 x. So we can see that this power here is not just x.

It's actually a whole function of x. It's x squared plus 4 x. 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. Here, we have this function inside of our exponential function.

So in finding this derivative here, g prime of x, we start by applying our rule for finding the derivative of an exponential function. So this is going to be 3x2+4x, 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 x2 plus 4 x, we know that using the power rule, this is 2x+4. And now this is my full derivative, g prime of 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.

2

Problem

Problem

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

$2\ln⁡x-5\ln x$

3

Problem

Problem

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

$4x^3+4+4^{x}\ln⁡4$

4

Problem

Problem

Find the derivative of the given function.
$y=(4x-3x^2+9)\cdot2^{5x}$

A

$2^{5x}\cdot[(4-6x)+5\ln⁡(2)(4x-3x^2+9)]$

B

$2^{5x}\cdot(-15x^2-14x+49)$

C

$2^{5x}[(4-6x)+\ln⁡(2)(4x-3x^2+9)]$

D

$5\ln⁡(2)(4-6x)\cdot2^{5x}$

5

Problem

Problem

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

$\left(\sqrt{x}+3x\right)^{\frac54}\cdot4\left(\sqrt{x}+3x\right)^{\frac14}$

6

example

Derivatives of General Exponential Functions Example 1

Video duration:

5m

Video transcript

In the previous video, we learned a rule for finding the derivative of our general exponential function, b to the x. Now a common exponential function that we'll encounter is e to the x. So how can we find that derivative? Well, because e to the 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 in order to then find the derivative of e to the 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. So in order to find the derivative of e to the 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 ex, 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 ex. This means that the derivative of ex is just ex. So what does that look like in action? Well, coming over here, we want to find the derivative of 4 times ex. Now using our constant multiple rule, I know that I can just pull that 4 out front, and this is 4 times the derivative of ex.

Now we just saw that the derivative of ex is just ex. So this is then 4 times ex, which is the exact same function that we started with. Now we're going to encounter some more complicated functions that involve e to the 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 to the 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 of x. This is 3 times e2x+4. So to find this derivative here, f prime of 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 e2x+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 in order to find this derivative. So when we start with that outside exponential function, ex, we know that the derivative of ex is ex. Since here, my exponent is not just x, that means that this derivative is e2x+4. Then applying the chain rule, we multiply this by the derivative of that inside function.

So the derivative of that 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 6 e2x+4 for my final derivative here, f prime of x. Let's work through one additional example. Here, we want to find the derivative of g of x, and this is x times e5x.

So here, I have a function x multiplying my exponential function e5x. That means 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 prime of 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, e5x. 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 going to be e5x, 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 to the 5x. Then continuing on with our product rule here, we add this together with right d left. So that's my right-hand function, e5x 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 e5x in them. I can go ahead and just factor that out. That becomes e5x times 5x. That's what I'm left over with here. Plus 1.

And this is my full derivative here, g prime of 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.

7

concept

Derivative of the Natural Exponential Function (e^x)

Video duration:

4m

Video transcript

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 in order 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 the derivative of \( e^x \).

Now we just saw that the derivative of \( e^x \) is just \( e^x \). So this is then 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 \( e \) to the power of \( 2x+4 \). 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 \( e \) to the power of \( 2x+4 \). 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 in order 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 \( e \) to the power of that whole interior function, that's \( 2x+4 \). Then applying the chain rule, we multiply this by the derivative of that inside function.

So the derivative of that \( 2x+4 \) 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 \( e \) to the power of \( 5x \).

So here, I have a function \( x \) multiplying my exponential function \( e \) to the power of \( 5x \). 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, \( e \) to the power of \( 5x \). 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 going to be \( e \) to the power of \( 5x \), 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 \( e \) to the \( 5x \). Then continuing on with our product rule here, we add this together with "right d left". So that's my right hand function, \( e \) to the power of \( 5x \) times the derivative of \( x \), which in this case is just 1.

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

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.

8

Problem

Problem

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

$-3e^x+5x-2$

9

Problem

Problem

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

$7e^x+2x^3$

10

example

Derivative of the Natural Exponential Function (e^x) Example 2

Video duration:

2m

Video transcript

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 \( x^3 \) or minus \( 2x \), 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 \( (x^3 - 2x)^{1/2} \). 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 \( 1/2 \). So the first thing I want to do is apply the power rule. So I pull that \( 1/2 \) out front, multiplying by it.

So \( 1/2 \times (x^3 - 2x) \), keeping that inside function the same, and then decreasing that power by 1. One half minus 1 gives me negative one half. 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 negative one half is really the square root on the bottom here.

So I can rewrite this as the square root of \( x^3 - 2x \). 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.

11

Problem

Problem

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

$x e^{3 x^{2} + 5} (6 x^{2} + 5 x + 2)$

12

Problem

Problem

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

$\frac{3t^2e^{3t-1}-8te^{2t-1}-e^{3t}}{(t-2e^{-t})^2}$

13

example

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

Video duration:

2m

Video transcript

Hey, everyone. In this problem, we want to find an equation to the tangent line of the function \( y = 3e^x - x \) 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, \( \frac{dy}{dx} \), our derivative of \( 3e^x - x \).

Since I have 2 functions here being subtracted, I can just subtract their individual derivatives. So the derivative of \( 3 \times e^x \) is just \( 3 \times e^x \). It's the same exact thing. And the derivative of negative x is just negative one. So my derivative here is \( 3e^x - 1 \).

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 \( 3 \times e^0 - 1 \). 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 \( y - y_1 = m \) (that's our slope) times \( x - x_1 \). Now we already have our slope here.

We know that it's just 2. We also know that \( x_1 \) is just 0. So the only thing that we have left to find here is \( y_1 \). 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 \( y_1 \), this is going to be \( 3 \times e^0 - 0 \), just plugging 0 in for \( x \).

So this just gives us 3 because subtracting 0 doesn't do anything. And \( e^0 \) is just 1. So now we can plug in \( y_1 \), \( x_1 \), and \( m \). So this is \( y - \) our \( y_1 \), which is 3, and that's equal to 2 times \( x - x_1 \), which is just 0.

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 \( y = 3e^x - x \) at \( x = 0 \). Let us know if you have any questions here and let's keep going.

14

concept

Derivatives of General Logarithmic Functions

Video duration:

6m

Video transcript

Hey, everyone. We just learned a rule for finding the derivative of our general exponential function, b to the power of x. Here, we're going to take a look at finding the derivative of our general logarithmic function. That's log base b of 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 equals log base b of x, my ultimate goal here is to determine my derivative, dydx. We wanna 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=by. 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 dydx. So having rewritten y=logbx as x=by, we now wanna 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 by, 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 by⁢lnb using my rule here. And then I wanna multiply this by the derivative of that inside function y since I know that that is a function of x. This is dydx.

Now this is ultimately the derivative that we wanna find. We wanna find the derivative of our logarithmic function. That's dydx. So we wanna solve for this dydx, which we can do by isolating it, dividing this over, this by⁢lnb, in order to get it to cancel. Then we're just left with that dydx.

Now swapping the sides of my equation here, this gives me that dydx=1byx⁢lnb. 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 base b of x.

So we can just plug that right in here. So this is blogbxlnb, just having replaced that y with this log base b of x. But using properties of logs and exponents, if I take b and raise it to the power of log base b, that just cancels out. So all I'm left with here is just x.

So then dy/dx, my derivative of log base b of x, is equal to 1x⁢lnb. 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 base b of x is 1x⁢lnb. 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 of x equals log base 8 of x. So applying my rule here in order to find a g prime of x, my derivative, this is then going to be equal to 1x⁢ln8.

And this is my full derivative. G prime of x is equal to 1x⁢ln8. Now we're gonna 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 of x.

This is log base 5 of x squared. So here, we can see that we're not just taking log base 5 of x. We're taking log base 5 of x squared. 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 prime of 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 base 5 of x squared, 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 a one over x squared, 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 squared.

So the derivative of x squared is just 2x. So we multiply this by 2x. That allows these x's to cancel, and I am left here with 2x⁢ln5 for my final derivative here, f prime of x, having applied both the chain rule and our new rule for finding the derivative of logarithmic functions. Now we're gonna continue getting practice putting all of our rules for derivatives all together. I'll see you in the next one.

15

Problem

Problem

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

$\frac{1}{x\ln8}$

16

Problem

Problem

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

$2+\frac{1}{t\ln5}$

17

example

Derivatives of General Logarithmic Functions Example 4

Video duration:

5m

Video transcript

Here, we're asked to find the derivative of the function \( y = \log_4 \left(\frac{\sqrt{x}}{4x+1}\right) \). Now from just taking a look at this function, we could see that there's quite a bit going on here. And 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 \(\log_4\) of \(x\), we're taking the \(\log_4\) of \(\frac{\sqrt{x}}{4x+1}\). 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 \(\log_4\).

Now we know that for a general logarithmic function, our derivative is going to be \(\frac{1}{x} \ln(\text{base})\). Since we're not taking the derivative of just \(x\) here, that makes this derivative \(\frac{1}{\frac{\sqrt{x}}{4x+1}} \ln(4)\). 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 square root of \( \frac{x}{4x+1} \).

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 \(\log_4\) term. Now we're working our way inside. We took the derivative of this square root of \(\frac{x}{4x+1}\). 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, \(\text{low } d(\text{high}) - \text{high } d(\text{low}) \text{ over the square of what's below}\). That's what's going to give us this derivative of \( \frac{x}{4x+1} \). So low \(d(\text{high})\), so my low function, \(4x+1\), times the derivative of that function on top. The derivative of \(x\) is just \(1\) minus high \(d(\text{low})\), 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 square root of \(\frac{x}{4x+1}\).

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 \( (4x+1)^2 \).

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 \( \frac{x}{4x+1} \) squared. So those square roots will cancel out and leave me with just one of those terms. So this is \( \frac{1}{2 \times \ln(4) \times \frac{x}{4x+1}}\) because those \(2\) multiplied together have just given me that inside term. Then finally, I have \(\times (4x+1)^2 \).

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 natural log of 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.

18

Problem

Problem

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

$\frac{x^3+2x}{x\ln5}+\log_5x\cdot\left(3x^2+2\right)$

19

Problem

Problem

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

$\frac{7^{t^{2+4}}}{t\ln5}$

20

concept

Derivative of the Natural Logarithmic Function

Video duration:

5m

Video transcript

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 over x times the natural log of e. But we know that the natural log of e is just equal to 1. So that leaves us here with just 1 over x.

And that's exactly what my derivative is. The derivative of the natural log of x is 1 over 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 over x, let's take a look at an example. Here, we want to find the derivative of 6 times the natural log of 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 times the derivative ddx of the natural log of x. Now we just found that the derivative of the natural log of x is just 1 over x. So this is then 6 times 1 over x, which we can rewrite a bit more compactly as just 6x. 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 gonna take a look at a slightly more complicated example here in finding the derivative of this function, f of x. Now this function is the natural log of x squared plus 4x. So we can see here that we're no longer just taking the natural log of x but the natural log of x squared plus 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 prime of x, we're gonna start with that outermost function, which here is the natural log. So we know that the derivative of the natural log is just 1 over 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 over x, but one over that entire function of x.

So in this case, that is a one over x squared plus 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 x squared plus 4x is 2x+4, and this is our full derivative. And we can write this a little bit more compactly with just 2x+4x2+4x. And this gives us our final derivative here, f prime of x, 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 wanna find the derivative of this function, g of x, which is x times the natural log of x cubed. So here, since I have this function x multiplying the natural log of x cubed, 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 prime of x here, I'm gonna 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 x cubed.

Now here, again, we can see that we're not just taking the natural log of x. We're taking the natural log of x cubed. 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 x cubed, I start with that outermost function. I know that the derivative of the natural log is a one over x.

But here, since I have x cubed, that makes this one over x cubed. And then I need to multiply by the derivative of that inner function x cubed, which is 3x squared. 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 x cubed 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 x cubed on the bottom is going to cancel out with that x times x squared on the top. That just leaves me with 3 plus the natural log of x cubed. So this is my full derivative at g prime of x. Now we can see that we put a lot of different derivative rules altogether in that example.

And we're gonna 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.

21

Problem

Problem

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

$6x-1$

22

Problem

Problem

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

$e^{x}+\frac{5}{x}$

23

Problem

Problem

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

$\frac{-3x^2-4x+3}{2\left(x+1\right)\left(x^2+3\right)}\cdot\ln\left(\frac{\sqrt{x+1}}{x^2+3}\right)$

24

Problem

Problem

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

$\frac{1}{x^2}$

25

Problem

Problem

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

$2e^{x^2\ln x}$

26

example

Derivative of the Natural Logarithmic Function Example 5

Video duration:

4m

Video transcript

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 for 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.

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 1,000 in for \( b \). Let's start with the derivative piece of this, so \( t'(b) \). Since I have this constant out front, I'm going to leave that out front.

That's \( 42 \times \) 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. Now that here is 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.

Now 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. Okay. When taking this derivative using the quotient rule, remember low \( d \) high minus 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 \( d \) low, so that's 650, times the derivative of \( 1500 - b \), which is just negative one, 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 \( 42 \times -(-1) \).

That's just positive one, so I'm left on the top with just 42 and then on the bottom with just \( 1500 - b \). So this is my derivative here. I found \( t'(b) \). Now we want to find \( t'(1000) \) by plugging 1,000 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. Now here, it can feel a little bit backwards because we took the derivative of time with respect to bacteria rather than with bacteria in respect to time. So we really calculated \( \frac{dt}{db} \). And 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 1,000 for \( b \). So this means that when our sample has 1,000 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 1,000 bacteria present, it is growing at a rate of 0.084 days per new bacteria. I hope that this was helpful, and let us know if you have any questions here. I'll see you in the next one.

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