Skip to main content

My Courses

3. Techniques of Differentiation

Basic Rules of Differentiation

3. Techniques of Differentiation

Basic Rules of Differentiation - Online Tutor, Practice Problems & Exam Prep

1

concept

Derivatives of Linear Functions

Video duration:

3m

Video transcript

So far, we've been learning to take derivatives using limits. So we could find the derivative of something like g(x)=7 or f(x)=x by setting up a limit and doing a bit of algebra to get to the answer. But what if I told you that there is a much quicker and easier way to find these derivatives, sort of a shortcut? In fact, you could look at these functions and know the derivative almost immediately by just learning a couple of rules. So throughout this chapter, I'm going to show you some rules for taking derivatives.

And while this may involve a bit of memorization, it will quickly become second nature. So here we're going to get started by taking a look at some rules for finding the derivatives of linear functions. No limits needed. So let's go ahead and dive into our example here. We're asked first to find the derivative of g(x)=7.

Now 7 is a constant and our first rule for finding derivatives tells us that the derivative, d/dx, of any constant, whether it's 7 or 12 or 1000000000, is always going to be 0. This makes sense, right? Because remember that the derivative is the slope of the tangent line to a function. So if I have a constant function like this one here, the slope of the tangent line to this function is always going to be 0.

So when asked here to find the derivative of 7, because 7 is a constant, I know my answer is just 0. Now let's take a look at the derivative of x. Now the derivative of x is always going to be equal to 1, and we can think about this in the same way. Remember that the derivative is the slope of the tangent line to a function. So if I have my function x here, the slope of the tangent line to this function at any point is always going to be 1.

So again here, the derivative, d/dx of x, is always going to be 1. Now we know that functions get more complicated than this. So what if I took these two functions and I added them together and I wanted to find the derivative of x+7? Well, this is where our sum and difference rule comes in. Our sum and difference rule tells us that when finding the derivative of one function f(x) plus or minus another function g(x), this is going to be equal to the derivative of that first function plus or minus the derivative of that second function.

So here, when finding the derivative of x+7, I know that I have two functions here. I have x and I have 7. So using my sum and difference rule, this is going to be equal to the derivative of x plus the derivative of 7. Now we know each of these individual derivatives based on what we saw above. I know that the derivative of x is just 1.

The derivative of 7 is 0, so I know that the derivative of x+7 is these two added together, just 1. Now let's add some multiplication into the mix. Here, we want to find the derivative of 8x. Now it's really important that you know that the derivative of 8x is not going to be the derivative of 8 multiplied by the derivative of x. Multiplication works a bit differently.

Now, because 8 is a constant, here we want to use our constant multiple rule, which tells us that whenever finding the derivative of a constant times a function f(x), we can just pull that constant out to the front and multiply it by the derivative of that function. So here, when finding the derivative of 8x, I want to take that 8, my constant, and pull it out front. So this is going to be equal to 8 times the derivative of x. Now, again, we know the derivative of x is just 1, so this is going to be equal to 8 times 1 or just 8. So here, we saw x+7 and 8x.

But you may be asked to find the derivative of something like 8x+7. And to do that, you can just use all of these rules together. As your functions get more and more complicated, you can just use multiple rules in order to find that derivative. And we're going to get some practice with that coming up next. I'll see you there.

2

Problem

Problem

Find the indicated derivative.
$f(x)=3x-4$
$f^{\prime}(x)=$

A

$0$

B

$1$

C

$2$

D

$3$

3

Problem

Problem

Find the indicated derivative.
$d/dt(-5t)$

A

$0$

B

$-5$

C

$5$

D

$5+t$

4

Problem

Problem

Find the indicated derivative.
$d/dt(-5x)$

A

$0$

B

$x$

C

$t$

D

$-5$

5

Problem

Problem

Find the indicated derivative.
$d/dx\left(\pi\right)$

A

$0$

B

$\pi$

C

$1$

D

$x$

6

concept

The Power Rule

Video duration:

4m

Video transcript

Hey, everyone. We just learned some rules that allow us to find the derivatives of some basic linear functions, like, say, x or 3x+4. But we know that functions can get more complicated than this. And often, we're going to have to deal with functions that have variables like x raised to a certain power. Like, here, we have x to the power of 6.

So we need to be able to find the derivative of functions like this, and luckily, we can do so using the power rule. Now the power rule is going to be massively important here as we continue to take more derivatives and as you continue on in calculus. So let's go ahead and jump right into what the power rule is. Now the power rule tells us that when taking the derivative of a power function xn, we can take that function and multiply it by that original exponent n. So, I can take that n and pull it out front, multiplying by it.

Then from there, I can decrease that exponent n by 1. So the derivative ddxxn is going to be equal to n×xn-1. So what does this look like for an actual power function? Well, let's come over here and find our derivative of x to the power of 6. Now using the power rule here, I want to take that original exponent 6 and pull it out front, multiplying by it.

Then I want to decrease that exponent by 1. So my derivative here, ddxx6, is going to be 6 times x to the power of 6 minus 1, or just 6x5. And we're all done here. Now we know that these power functions will not always be all by themselves like this x to the power of 6 is here. Often, we're going to be dealing with some addition or subtraction or maybe even some multiplication.

So let's go ahead and take a look at a couple more examples and apply that power rule alongside the rules that we've already learned. So looking at this first function here, I have f(x) is equal to x to the power of 4 plus x cubed. So since I have 2 functions being added together here, in order to find my derivative f'(x), I want to split this up using the sum rule. So I want to find my derivative at ddxx4 plus ddxx3. And now I can apply the power rule to each of these functions separately.

So starting here with x to the power of 4 using my power rule, I want to pull that 4 out front multiplying by it and then decrease that exponent by 1. So that gives me here a 4 times x to the power of 4 minus 1, which is 3. Then for that second term, finding my derivative again here using the power rule, I want to pull that 3 out front multiplying by it and then decrease that exponent by 1. So this gives me here a 3 times x to the power of 3 minus 1, which is 2. So my final answer here for my derivative is going to be a 4x cubed plus 3x squared having used the power rule and the sum rule together.

So let's go ahead and look at one final example here. Here we have our function g(x) is equal to 3x squared. Now here I have a constant 3 multiplying this function x squared. So using my constant multiple rule to find my derivative, I can find a g'(x). Pulling that 3 out front, this is going to be 3 times ddxx2.

Now I can go ahead and apply my power rule to this power function here x squared. So keeping that 3 out front as it is, multiplying that by my derivative here using the power rule, pulling that 2 out front, multiplying by it, then decreasing that exponent by 1. This gives me 2 times x to the power of 21, which is just 1. Then I can simplify this going ahead and multiplying ese 3 times 2 x. That gives me a 6 x for my final derivative here, a g'(x).

So you now know one of the most important rules for taking derivatives, the power rule. And as we continue to practice, you're going to get even quicker and better at taking these derivatives. So let's continue on this journey together. I'll see you in the next video.

7

concept

The Power Rule: Negative & Rational Exponents

Video duration:

4m

Video transcript

Hey, everyone. We just learned the power rule, which is going to be massively important as we continue to take derivatives. But what if I asked you right now to find the derivative of x-3, or maybe even x3⁄2? Now looking at these functions, it's clear that they are power functions, but their exponents look a little bit different. So you may be wondering, can I still use the power rule, or do I have to learn a completely new rule to take these derivatives?

Well, the good news is that you can still use the power rule because the power rule works for any power function, including those with negative or rational exponents, just like we see in these functions here. So all we're going to do here is to keep using the power rule and specifically apply it to some examples, including negative and rational exponents. So let's go ahead and jump right into our first example here where we want to find the derivative of f(x)=x-3. Now even though that power is negative, it doesn't matter. I can still use my power rule to find my derivative here f′(x) and applying my power rule here, pulling that exponent out to the front, multiplying by it, and then decreasing that exponent by 1.

This gives me a negative 3 times x-4. Now this is already a correct answer but we could also choose to rewrite this without that negative exponent by simply pulling that term to the bottom. So this is -3x4. Now both of these are correct answers and we are all done here. Now let's move on to our next example.

Here we're asked to find the derivative of x3⁄2. Now again here even though we have a rational exponent it does not matter. We can still use our power rule. So using our power rule to find our derivative here f′(x) we're pulling that power out to the front, multiplying by it and then subtracting 1. So this gives me 3 halves times x1⁄2. Now let's keep this going and move on to our next example. Here we have f(x)=1x. Now looking at this, this does not look like something like xn. This does not look like a power function.

But that's okay because sometimes you may actually need to rewrite a function as a power function before taking the derivative. And that's exactly what we want to do here. Now this is already something that we used back in example a when we rewrote that x-4, remember that whenever we have a one over x, we can rewrite this as x-1. Now we can clearly use the power rule the same way that we know how. So continuing on here finding our derivative f′(x), we're pulling that negative one out to the front and then decreasing that power by 1 using our power rule.

This gives us negative one times x-2. Now again, we can rewrite this without that negative exponent putting it back in a similar form to our original function by rewriting this as -1x2 for our final derivative here. Now let's move on to our final example here. We have f(x)=x3. Now, again, looking at this function, it may not be clear how we're going to use the power rule.

But remember, you may have to rewrite the function as a power function before you take the derivative, which is exactly what we're going to do here. I have the cube root of x, which I can rewrite as x1⁄3 ending up with a rational exponent there. Now I can go ahead and use the power rule. Now I'm using the power rule here to find our derivative f′(x). We are, of course, multiplying by that exponent and then decreasing it by 1.

So this gives us 1 third times x-2⁄3, and this is our final answer here. So now that we have a better idea of the power that the power rule has to take derivatives, let's continue getting some practice. Thanks for watching, and I'll see you in the next video.

8

example

The Power Rule: Negative & Rational Exponents Example 1

Video duration:

3m

Video transcript

In this problem, we're asked to find the derivative of the given function, and the function that we're given here is \( y = 2x^3 - x^2 + 4x + 7 \). Now I'm going to fully break down this derivative with all of the rules for derivatives that we know. Now as you get better and better at taking derivatives, you're not going to have to break it down fully in this way. But for now, I'm going to show you the entire process of taking this derivative. So let's go ahead and get started.

Now here we have our function \( y \). So our derivative here, we could denote as \( \frac{dy}{dx} \) because we want to take the derivative with respect to this variable \( x \). Now you could also denote this derivative as \( y' \), and that would still be accurate. Now taking our derivative here, \( \frac{dy}{dx} \), I have a bunch of different terms being added together, and I also have some multiplication of some constants happening. So in order to fully break this down using all of the rules for derivatives that we know, we're going to go ahead and separate all of these out.

Now we know that whenever we have a constant, we can already pull that out to the front of our derivative. So when thinking about this first term here, \( 2x^3 \), I can think of this derivative as \( 2 \times \frac{d}{dx} of x^3 \), and I'll eventually take that derivative using the power rule. Now for my next term, I have a minus \( x^2 \). So here, using my sum and difference rule, specifically here a difference, I am subtracting here the derivative \( \frac{d}{dx} \) of that \( x^2 \). Then for my next term, I have plus \( 4x \).

Since I have another constant, I can think of this as pulling that 4 out to the front of my derivative, and this is plus \( 4 \times \frac{d}{dx} \) of \( x \). Then lastly, I have a constant down here. So I have plus \( \frac{d}{dx} \) of 7. Now I can think of all of these derivatives individually and add and subtract them as needed. Now here, looking at this first derivative here, \( 2 \times \frac{d}{dx} \) of \( x^3 \).

I already have that 2. I want to keep that as is, that constant. And then using the power rule here, I want to go ahead and pull that 3 out to the front and then decrease that power by 1. That's using the power rule that we just learned. So this becomes \( 3 \times x^{3-1} \), which is just 2.

Now from here, we can take our next derivative. Here, I'm subtracting this derivative, remember? And again, I'm going to be using the power rule here. Now with this \( x^2 \) I'm going to pull that 2 out to the front and then decrease that exponent, that power, by 1. So this becomes \( 2 \times x^{2-1} \), which is just 1 so we don't have to worry about writing it.

Then adding my next derivative, I have plus \( 4 \times \frac{d}{dx} \) of \( x \). Now we know that the derivative of \( x \) is just 1, so this is really just 4 times 1. Then lastly, I have the derivative of my constant down here, \( \frac{d}{dx} \) of 7. I'm adding that together here. Now the derivative of any constant by itself is just 0, so I'm just adding a 0 on there.

Now we can do some more simplification here by just multiplying some stuff out and taking away that 0 on the end. So multiplying this term out, \( 2 \times 3x^2 \), it gives me \( 6x^2 \), and then I have a minus \( 2x \) that next term, plus \( 4 \times 1 \), which is just 4. And this gives me my final derivative \( \frac{dy}{dx} \) of that original function. Now remember, like I said, as you continue taking more and more derivatives, you're really going to be able to skip right from your original function to your derivative without having to write out all of those steps. Now as we get there, we're going to continue to get more practice. So feel free to let me know if you have any questions, and I'll see you in the next one.

9

Problem

Problem

Find the indicated derivative.
$f(x)=6x^4-4^3$
$f^{\prime}\left(x\right)=$

A

$4x^3$

B

$24x^3$

C

$4x^3-1$

D

$24x^3-1$

10

Problem

Problem

Find the indicated derivative.
$d/dt(5-7t^{-3})$

A

$1-21t^{-4}$

B

$-21t^{-2}$

C

$-3t^{-4}$

D

$21t^{-4}$

11

Problem

Problem

Find the indicated derivative.
$y=\frac52x^5+2x^3+6x-\frac{\surd3}{4}$
$y^{\prime}=$

A

$5x^4+3x^2+6$

B

$\frac{25}{2}x^4+6x^2+5$

C

$\frac{25}{2}x^5+6x^3+6x-1$

D

$\frac{25}{2}x^4+6x^2+6$

12

Problem

Problem

Find the indicated derivative.
$h(t)=\frac12t^3+\frac{4}{t^2}t+3\sqrt{t}$
$h^{\prime}(t)=$

A

$t^2-\frac{2}{t^3}+\frac{1}{\sqrt{t}}$

B

$\frac32t^2-8t^3+\frac32\sqrt{t}$

C

$\frac32t^2-\frac{8}{t^3}+\frac{3}{2\sqrt{t}}$

D

$\frac32t^3-\frac{8}{t^2}+\frac32\sqrt{t}$

13

example

The Power Rule: Negative & Rational Exponents Example 2

Video duration:

5m

Video transcript

Hey, everyone. In this problem, we're asked to answer the questions below given the function f of x equals 2x3-3x+5. So let's dig into these questions. Our first question a asks us to find f prime of x, the derivative of our function f of x. I'm going to go ahead and do my work for this down here.

So here, starting with part a, I want to find the derivative f prime of x. Now looking at my original function, I know that I can go ahead and use the power rule on this polynomial. Now using the power rule here, I want to pull that 3 out to the front, decrease that exponent by 1. Now that gives me here 6x2-3. The derivative of any constant is 0. So I already have my final answer here.

The derivative f prime of x is 6x2-3. So we're done with part a. Let's go ahead and move on to part b. Now in part b, we're asked to find the equation of the tangent line at x equals 1. So what information do we need to be able to do this?

Well, I know that I can set up my point slope formula, y minus y₁ equals m times (x minus x₁). Based on this point slope formula, what is all of the information that we need to find the equation of this tangent line? Well, I need y₁, I need m, and I need x₁. So what do we need to find here? Well, I already know that x₁ is going to be one because that's already given to me here.

This is my x₁ value. Now I need to find m, which is my slope, and I need to find y₁, which is just the value of my original function with x₁ plugged in. So let's go ahead and start there with y₁. Now in order to find y₁, we just need to plug 1 into our original function. So this is f of 1.

Now plugging that into our original function, which is that 2x3-3x+5, I have 2⋅13-3⋅1+5. Now this ends up giving me a positive 4. So this is my y₁ value that is ready to plug into that point slope formula. But we need to find one more thing here. That's our m, our slope value.

Remember that the derivative of a function is the slope of its tangent line. So in order to find the slope of our tangent line specifically at x equals 1, we just need to plug 1 into our derivative. Now we already found our derivative in part a, so in order to find m here I'm just going to find f prime of 1 plugging in that x value that we're given. So f prime of 1, having found my derivative in part a, this is 6⋅12-3. Now 6⋅12 is just 6, so this is 6 - 3, which is just 3, and this is my m value.

Now we can plug all of this into our point slope formula. So scrolling down a little bit here since I need a little bit of space, plugging all these values in, I have y minus y₁, which here is 4. So plugging that 4 in, y minus 4 is equal to m, my slope, which we found was 3. So 3 times (x minus x₁), which again was already given to us. That was our value of 1 given in our problem statement.

We can simplify this because we don't typically present equations of lines in this way. So I'm going to go ahead and distribute this 3 into these parentheses and also move this 4 over to this side. So this becomes y equals 3x-3, distributing that 3, plus 4. Now -3+4 is just positive 1. So my final answer here for the equation of my tangent line at x equals 1 is y equals 3x+1.

Now we've completed part b. We have one final question here, part c, which asks us to find the values of x for which the tangent line is horizontal. Now how exactly can we figure this out? What does that even mean the tangent line is horizontal? Well, when we think of a horizontal line remember that the slope of a horizontal line is 0.

Now, we know that the slope of our tangent line is our derivative. Since we found our derivative we can take that derivative f prime of x and set it equal to 0 in order to find all of those x values for which our tangent line is horizontal. So let's go ahead and plug this in. I have 6x2-3. That's my derivative here that I am setting equal to 0.

Now if I actually solve here, I can add 3 to both sides. Canceling it out here, I have 6x2=3. Dividing by 6 on both sides gives me x² equals 36, which is just 1/2. Then my final step here is to take the square root. So my final answer here is x equals plus or minus the square root of 1/2.

Remember, since we're taking the square root we want to account for all of those values positive or negative. Now the square root of 1/2 you may also see this written as root 2 over 2, so this is also an acceptable answer, plus or minus root 2 over 2 or plus or minus the square root of 1/2. Now we're finally all done here. I know this was a rather tedious problem, but we want to get as much practice as possible here. Thanks for watching, and let me know if you have any questions.

14

example

The Power Rule: Negative & Rational Exponents Example 3

Video duration:

3m

Video transcript

Now here, we're given an application problem, meaning that we're going to have to use our skills for derivatives but in the context of a real-world example. This is something that you'll be asked to do from time to time, and it's really important to be able to think critically about derivatives in the context of some scenario that's actually happening. So let's go ahead and dive into this example. Here, we're told that the number of cells in a cell culture is given by the function pt, where p is the number of cells and t is the time measured in hours. Now we want to find p24 and p'24, my derivative, also at 24.

Now we also want to interpret what this actually means and what this tells us about the cell culture. Let's dive in here. I want to start by finding p24 and determining what that tells us about our cell culture. p24 tells us that we just need to plug 24 into our original function here, which is p(t)=1500+2t2. I'm going to plug 24 right in here.

p24=1500+2×242. After calculating, this gives me 2,652. What does this value actually mean in the context of this problem? p is the number of cells after a certain amount of time, t, has passed.

If we plug in 24 here, we know that time is measured in hours. This is after 24 hours have passed. Since p(t) is the number of cells, this tells us that after 24 hours, there are 2,652 cells present in our cell culture. Now that we have determined that p24 and what that means, let's move on to our next task, which is finding our derivative p'24. To find p'24, we first need to find our derivative p'(t).

Using the power rule, the derivative of any constant is just 0, and for that second term, applying the power rule becomes 4t. Now that we have our derivative, I can plug in the 24 as asked in the problem. So p'24=4×24. This computes to 96, which is my value p'24. But again, we want to think critically about what this actually means.

The derivative is a rate of change. If p is the number of cells, then p', my derivative, is the rate of change of the number of cells. Given 24 hours, this tells me that after 24 hours, the rate of change of my cells is 96 cells per hour. This challenges us to think critically about what derivatives mean in the context of something else.

Thanks for watching. Let me know if you have questions.

15

example

The Power Rule: Negative & Rational Exponents Example 4

Video duration:

3m

Video transcript

In this problem, we're told that a business's total sales in thousands of dollars \( t \) months from now are given by the function below: \( s(t) = 2.4t^3 + 0.4t^2 + 5t - 1 \). We are asked to find \( s(6) \) and \( s'(6) \), our derivative, also at 6, and to also consider what these values tell us about the business.

First, we want to find \( s(6) \), which means plugging 6 into our function. This function indicates the total sales of the business after a specific number of months. Substituting 6, we have: \( s(6) = 2.4 \cdot 6^3 + 0.4 \cdot 6^2 + 5 \cdot 6 - 1 \). Calculating this gives \( s(6) = 561.8 \). Given this is in thousands of dollars, this equates to $561,800. This informs us that after 6 months, the business has made $561,800 in total sales.

Next, we need to determine \( s'(6) \), which is the derivative evaluated at \( t = 6 \). The derivative \( s'(t) \) of the function by using the power rule is calculated as follows: \[ s'(t) = 7.2t^2 + 0.8t + 5 \] With \( s'(6) = 7.2 \cdot 6^2 + 0.8 \cdot 6 + 5 \), which simplifies to \( s'(6) = 269 \). When interpreting this, considering that the derivative represents the rate of change, we conclude that 6 months after establishment, the rate of change of the business's sales is $269,000 per month.

Please let me know if you have any questions regarding this problem. These concepts can sometimes be challenging to grasp. I'll see you in the next video.

Your Business Calculus tutors

Math Instructor

Math Instructor