Product and Quotient Rules - Online Tutor, Practice Problems & Exam Prep

The Product Rule

We know that in order to find the derivative of two functions being added together, we can take their derivatives separately and then add them together. But what about two functions being multiplied? Like here, we have our function h(x), which is (x - 5) times (2x + 9). Now, based on what we know about finding the derivative of two functions being added, you may think here that we can take the derivative of each of these functions separately and then multiply them together. But that's actually not the case at all.

See, the derivative of two functions being multiplied is not going to be equal to their derivatives multiplied. Instead, we have to use what's called the product rule. So here, I'm going to walk you through exactly what the product rule is and how to use it, and we'll work through some examples together. So let's go ahead and get started. Now coming back down here to our function h(x), we have these two functions being multiplied: (x - 5) and (2x + 9). We could go ahead and just foil this out and find the derivative using all the rules that we've already learned, but it's actually much quicker and easier to know and use the product rule.

So let's go ahead and look at what the product rule even is. The product rule tells us that when finding the derivative of two functions being multiplied together, that's f(x) times g(x), we can take that original first function, f(x), just leaving it as is, and multiply it by the derivative of that second function, g'(x).

Then we can add that together with that second function as is, g(x), multiplying it by the derivative of that first function, f'(x). Now depending on your professor or your textbook, you may see the product rule written in a slightly different order, but it's still the exact same thing. It's going to be really useful to go ahead and memorize the product rule, and I have a tool to help you do so. If we think about f(x) and g(x) regarding their positions here, we have f(x) on the left and g(x) on the right; we can think about finding their derivative using the product rule by remembering this mnemonic device: "left d right plus right d left," where "d" just means we're taking the derivative.

Now looking at each of the two terms in our product rule, we can see that each of them contains one original function times the derivative of the other function. So there are a couple of different ways that you can remember the product rule. Now let's go ahead and take a look at the product rule in action, coming back over here to our function h(x). Now we have these two functions, (x - 5) and (2x + 9). And I'm going to think to myself when taking the derivative "left d right plus right d left."

So starting with that left-hand function, that's (x - 5), leaving that as is, that's an original function, multiplying that by the derivative of that right-hand function, (2x + 9). Now the derivative of (2x + 9) using my power rule is just going to be 2, and then I'm adding that together with that right-hand function as is (2x + 9) times the derivative of my left-hand function (x - 5). The derivative of this function is just 1.

Now, from here, I have fully applied the product rule, and all we have to do is simplify. So let's go ahead and distribute these constants here. Distributing that 2, distributing that 1, this first term, having multiplied both of these by 2, gives me 2x - 10. Then that second term, since we're just multiplying by 1 here, this remains 2x + 9. Now we can combine some like terms here. I have 2x and 2x. That gives me 4x.

And then I have a minus 10 + 9, which gives me minus 1 for my derivative here, h'(x), and we are all done. Now, we want to get really good at using the product rule, so let's take a look at one more example here. Here, we want to find the derivative of this function, y equals (2x^2 - 1) times (3 + x^3). Now, to find our derivative here, y', I'm thinking to myself "left d right plus right d left." So I have my two functions: that function on the left, (2x^2 - 1), and that function on the right, (3 + x^3).

So starting with that left function, that first term is "left d right." So that left function, (2x^2 - 1) times the derivative of my right function, that's (3 + x^3). The derivative of this function using the power rule is going to be 3x^2. Then I'm adding this together with that second term here, "right d left," taking that right-hand function, leaving it as is, (3 + x^3), and multiplying it by the derivative of that left-hand function.

Again, using the power rule here, I get the derivative as 4x. So distributing that 3x^2 into that (2x^2 - 1) is going to give me 6x^4 - 3x^2. Then distributing this 4x into this (3 + x^3) is going to give me 12x and then plus 4x^4. And we have some like terms we can combine here. 6x^4 and 4x^4 give me 10x^4 and then a minus 3x^2 plus 12x.

This gives me my final derivative here, y'. Now, we know what the product rule is and how exactly to use it for two functions that are being multiplied. Now we want to get tons of practice with this product rule, so I will see you in that next video.

The Product Rule Example 1

Hey everyone. In this problem, we're asked to find the derivative of this function \( y = 4x^2 \times (8 - x^3) \) using 2 different methods. The first way we want to find this derivative is using the product rule that we just learned. And the second way is by fully multiplying out our expression and simply using the power rule. So let's go ahead and get started here, starting by finding this derivative using the product rule. Now using the product rule here, remember that we consider these two functions being multiplied together and we look at them as our left side function and our right side function.

So we have our function on the left, this \( 4x^2 \), and our function on the right, \( 8 - x^3 \). Now from here, we actually want to apply our product rule. Remember, our product rule tells us that our derivative here, \( y' \), is going to be equal to left \( d \) right plus right \( d \) left. Whenever I say \( d \), that just means we're taking the derivative. So that very first term, left \( d \) right, I'm going to take that left side function, which is \( 4x^2 \), leaving that as is.

And I'm going to multiply that by the derivative of that right side function. Now, the derivative of \( 8 - x^3 \) using the power rule, if I pull that 3 out to the front and decrease that power by 1, now I know that the derivative of a constant is just 0. That's why I'm not worrying about it here. This derivative is going to be negative 3 times \( x \) to the power of \( 3 - 1 \), which is equal to 2. Now that's our first term here.

Our second term is that right side function. So \( 8 - x^3 \), leaving that as is times the derivative of that left-hand function. So multiplying by that here, the derivative of this left side function, again, using the power rule, pulling this 2 out to the front, multiplying by it, decreasing that power by 1. 2 times 4 using that power rule is going to be 8 and then times \( x \) to the power of \( 2 - 1 \), which is just a one, so I don't have to worry about writing an exponent there. Now from here, we can multiply this out and do some simplifying.

Now if I simplify here, \( y' \) is going to be equal to multiplying this out. This is going to be a negative 12\( x^4 \). Then for that second term, I have a plus, and then I'm going to go ahead and distribute this \( 8x \) into my parentheses here. This is going to give me \( 64x - 8x^4 \). Now we can do some further simplification here because I have some like terms, negative 12\( x^4 \) and negative 8\( x^4 \).

This is going to give me negative 20\( x^4 + 64x \) for my final derivative here, \( y' \). Now, we're not done yet because we found this derivative using our first method, the product rule. But we also want to go ahead and try this out by multiplying out the expression and using the power rule. So let's go ahead and try this second method here by multiplying our function out. Remember, our function here is equal to \( 4x^2 \times (8 - x^3) \).

So let's go ahead and distribute this \( 4x^2 \) into our parentheses here multiplying. This makes our function equal to 8 times \( 4x^2 \), which is \( 32x^2 \), and then minus \( 4x^5 \). Now, actually, using the power rule here now, since we have this function fully multiplied out. We no longer have 2 functions multiplying each other. I can go ahead and apply my power rule to find a \( y' \), my derivative.

Now in order to find this derivative using the power rule, I'm pulling that 2 out to the front, multiplying by it, decreasing that power by 1, giving me a first term here of \( 64x \) to the power of \( 2 - 1 \), which is just 1. Then for my second term here, again, using the power rule, pulling that 5 out to the front, multiplying by it, decreasing that exponent by 1, this is going to give me a negative 20\( x^4 \). So this is my derivative using that second method. Now we can see that we got the same exact answer for these two derivatives no matter what method we use. More often, it's going to be much more simple and quick to use the product rule.

But remember that if you ever get tripped up and you want to double-check yourself, you can always just multiply out your expression and use the power rule, which I'm sure you are super great at using by now. Now if you have any questions, feel free to let me know, and let's get some more practice with the product rule.

Find the derivative of each function.

$y=(3x+5)^2$

• A. $3$
• B. $6x$
• C. $9x+15$
• D. $18x+30$

Find the derivative of each function.

$f\left(t\right)=2t(t^{-3}+t^{2/3})$

• A. $2t^{-3}+2t^{2/3}$
• B. $-3t^{-4}+\frac23t^{-1/3}$
• C. $-4t^{-3}+\frac{10}{3}t^{2/3}$
• D. $-6t^{-3}+\frac{4}{3}{t}^{2/3}$

The Product Rule Example 2

Hey, everyone. In this problem, we're asked to find the slope of the tangent line to the curve y=x2-5×(6x+1) at a value of x=2. Now it may take you a second to recall exactly how to do this, but remember that the slope of a tangent line is just equal to the derivative. So, what we need to do first here is find the derivative. So looking at our function, y=x2-5×(6x+1), I see I have two functions multiplying each other, x2-5 and 6x+1.

So, to find the derivative of y′, I can use the product rule. So finding y′ here using that product rule, remember the product rule tells us left d right plus a right d left. That is our derivative. We're gonna start with that left-hand function leaving it as is x2-5 and we're gonna multiply that by the derivative of that right function 6x+1. Now, the derivative of 6x+1 is just going to be 6.

So that's what I'm multiplying by here. Then I can move on to that next term in my product rule, which I'm going to take that right-hand function 6x+1, leaving it as is, and then multiply it by the derivative of my left function, x2-5. Now, the derivative of this function using the power rule, I'm multiplying by that 2, decreasing that power by 1. That is going to be 2x. I don't have to worry about that negative 5 because the derivative of any constant is just 0.

Now from here, I can do some simplification by going ahead and distributing this 6 in these parentheses here as well as this 2x. Now doing that here, that gives me 6x2-30 and then a plus 2x times 6x is going to give me 12x2. And then a plus one times 2x is just 2x. We can do a bit more simplification here because we have some like terms. We have 6x2 and 12x2.

That's going to give us 18x2. And then rewriting this just in its standard form, we're going to write plus 2x next and then minus our constant, 30. And this is our derivative here, y′. But we are not quite done yet because remember, we want to find the slope of the tangent line at a specific value, x=2. So we need to go ahead and plug 2 into this function.

So to actually get this value here, we're gonna have 18×22+2×2-30. Now actually solving this out here, 18×4, which is 72, plus 2×2, which is 4, and then minus 30. 72 plus 4 is 76 minus 30 gives us 46. So our final answer here is 46. This is the slope of the tangent line to our curve, that original function we had, at a value of x=2.

Feel free to let me know if you have any questions here, and definitely try to get some more practice with these problems. I'll see you in the next one.

The Quotient Rule

We just saw that when finding the derivative of two functions being multiplied, we could not just multiply their derivatives. We had to use the product rule. So what about two functions being divided? Like here, we have the function h of x, which is x over 3x minus 4. Now, similarly, to two functions being multiplied, the derivative of two functions being divided is not just equal to their individual derivatives divided.

Instead, we have to use the quotient rule. So here I'm going to walk you through what the quotient rule is, and we'll work through some examples together. So let's go ahead and jump in here. Now, coming back down to our function h of x, we have two functions. We have our function x, and that is over our second function, 3x minus 4.

To find this derivative, we need to use the quotient rule. Now the quotient rule tells us that when finding the derivative d/dx of one function f of x divided by another function g of x, we take that function on the bottom, g of x, and multiply it by the derivative of that top function, f'(x). Then from here, we subtract that top function f of x multiplied by the derivative of that bottom function, g'(x). Then all of that gets divided by that bottom function, g of x, squared. Now I know that the quotient rule looks rather complicated.

So I want to give you a tool to help you memorize it. Now if we think about these two functions, f of x and g of x, in respect to their positions here. We have f of x on the top. We're going to refer to this as our high function. And we have g of x on the bottom. We're going to refer to that as our low function.

We can remember the quotient rule using this mnemonic device, saying to ourselves, "low d high minus high d low over the square of what's below." We can say this rhyme to ourselves any time we need to find the derivative using the quotient rule. When we say "d" here in this rhyme, that just means we're taking the derivative. So let's come back over here to our function h of x and actually apply our quotient rule here. Now I have my function on the top x, my function on the bottom, 3x minus 4, and I'm going to say to myself, okay. "Low d high minus high d low."

That's going to be my numerator here. So I want to take that low function, my function on the bottom, 3x minus 4, just leaving that function as is, and multiply it by d high. That's the derivative of that function on the top. Now the derivative of x is just 1, so that's 3x minus 4 times 1. Then here, I am subtracting high d low, so that's that high function on the top x, times the derivative of that function on the bottom.

Now the derivative of 3x minus 4 is just 3, so that's what I'm multiplying by here. Then this is all over the square of what's below. And so I'm going to put that 3x minus 4 on the bottom and square that. Now from here, we have applied our quotient rule, and we just need to simplify. So let's go ahead and simplify this numerator here by distributing this one into this 3x minus 4 and just cleaning this up.

Now 3x minus 4 times 1 is just 3x minus 4. Nothing is going to change there. And then we are subtracting. Multiplying x times 3 gives me 3x. And all of this is over that 3x minus 4 squared.

Now I see I have these like terms that are actually going to cancel. 3x minus 3x cancels out, and this just leaves me here with negative 4 over 3x minus 4 squared for that final derivative here, h' of x. Now as we've seen throughout the past couple of videos, we know that the most important part of learning all of these rules is getting a ton of practice with them. So let's go ahead and work through one more example together here, again, using our quotient rule. So here we want to find the derivative of y equals 2x squared minus 1 over 3 minus x cubed.

So I have that function on the top, 2x squared minus 1, over this function on the bottom, 3 minus x cubed. And I'm going to say to myself, okay, "Low d high minus high d low over the square of what's below." That's what's going to give me my derivative here, y'. So setting that up, I know that in my numerator, my first term here is that low d high.

So I'm taking that bottom function, 3 minus x cubed, that low function, and multiplying it by the derivative of that top function. That's that d high. Now the derivative of 2x squared minus 1 is just going to give me a 4x. And now I'm subtracting that second term, high d low, leaving that top function as is 2x squared minus 1, and then multiplying that by the derivative of that low function on the bottom there. Now the derivative of 3 minus x cubed is going to be negative 3x squared.

Now we know that all of this is over the square of what's below here. Now on the bottom, I have that 3 minus x cubed, and it's over the square of what's below. So I need to make sure that I square that. Now from here, we have applied our quotient rule, and it's just algebra and simplifica

Find the derivative of the function.

$y=\frac{2-3t}{4t^2+7}$

• A. $\frac{12t^2-16t-21}{(4t^2+7{)}^2}$
• B. $\frac{-12t^2+16t+21}{(4t^2+7{)}^2}$
• C. $\frac{-12t^2+16t+21}{(2-3t{)}^2}$
• D. $\frac{12t^2-16t-21}{(2-3t{)}^2}$

Find the derivative of the function.

$f\left(x\right)=\frac{2x-1}{x^3+2}$

• A. $\frac{4x^3-3x^2-4}{(x^3+2{)}^2}$
• B. $\frac{4x^3-3x^2-4}{(2x-1{)}^2}$
• C. $\frac{-4x^3+3x^2+4}{(x^3+2{)}^2}$
• D. $\frac{8x^3+3x^2+4}{(x^3+2{)}^2}$

Find the derivative of the function.

$g\left(y\right)=\frac{5y^2+2y-1}{y^2-2}$

• A. $\frac{2y^2+18y+4}{(y^2-2{)}^2}$
• B. $\frac{-2y^2-18y-4}{(y^2-2)^2}$
• C. $\frac{2y^2+18y+4}{(5y^2+2y-1)^2}$
• D. $\frac{-2y^2-18y}{(y^2-2)^2}$

The Quotient Rule Example 3

In this problem, we're asked to find the derivative of the function y equals (x+1)(x²-3x) over x³. Now you should go ahead if you haven't already and try to solve this problem on your own because you already have all of the skills that you need to do so. But I'm going to go ahead and jump into finding this derivative here, a dy/dx of my function. Now the first thing that I notice about my function here is that I have one function, this function on the top, being divided by another function on the bottom here, meaning I need to use the quotient rule. Now remember the quotient rule tells us "low d high minus high d low over the square of what's below."

So going ahead and applying that quotient rule here, I start with my low function, x³, and multiply it by the derivative of that top function, my high function. Now looking at this function on the top, I have (x + 1) multiplying (x² - 3x). So in order to take this derivative, I need to use the product rule within the quotient rule that I'm already using here. But we know how to do this. So let's go ahead and focus on finding this derivative of this top function here.

Now I have two functions being multiplied, that function on the left and my function on the right. Remember, our product rule tells us left d right plus right d left. That's our derivative. So I'm going to take that left-hand function first, x + 1, and multiply it by the derivative of this right-hand function using the power rule that gives me 2x - 3. Then I am adding this together with my right-hand function, x² - 3x, leaving that as is, times the derivative of that left-hand function.

The derivative of x + 1 is just one, a much easier derivative there, and that's that full derivative of that top function. So we can move on in our quotient rule here. We are subtracting from this low d high. We're subtracting high d low. So taking that high function as is, which is rather long, (x+1)(x²-3x) and multiplying that by the derivative of that low function, a little bit easier to find here using the power rule, that is 3x².

So that is my full numerator here. I'm dividing this all by the square of what's below, so that is x³ squared. And now from here, we have applied our quotient rule fully, and we just have a bunch of algebra to follow. Now because we have so much going on in our numerator because we had to use the product rule, we need to be super careful here when doing all of this algebra. So make sure that you stay organized, take your time so that you can get the correct answer here.

So let's continue on doing some algebra. Now the first thing that I'm going to do is I'm going to look at everything that I need to multiply out and focus on each of those terms. Foiling out and combining like terms carefully, we end up with a simplified polynomial. Remember, our denominator is still x to the power of 6.

Finally, we distribute negatives, combine like terms one last time, and simplify by canceling out powers of x where possible, as all terms involve x. After all the cancellations and simplifications, we arrive at dy/dx = (2x + 6) ÷ x³, which simplifies further to dy/dx = (2x + 6)/x³. I know that this is a long process, but this is our final derivative here for dy/dx of our original function.

It is (2x + 6)/x³. If you made it this far in the video, thanks so much for watching, and of course, feel free to let us know if you have questions.

The Quotient Rule Example 4

In this problem, we're given quite a bit of information. We're told that \( f(2) \) equals 4. We are also informed that \( f'(2) \), the derivative, is equal to negative 1, and we have information about a second function. We're informed that \( g(2) \) is equal to 3 and \( g'(2) \), the derivative of that function, is equal to 0. Now, we are asked to find \( h'(2) \) but for two different scenarios: when \( h(x) = f(x) \times g(x) \) as well as when \( h(x) = \frac{f(x)}{g(x)} \).

This sort of problem might seem intimidating because it appears that there's a lot of information being thrown at you. But when you break it down into what's actually being asked, you know exactly how to do this problem, and we can get through it rather quickly. So let's go ahead and jump in here. Part A asks us to find \( h'(2) \) when \( h(x) = f(x) \times g(x). \)

If that function, \( h(x) \), is equal to \( f(x) \times g(x) \), and we want to find \( h'(2) \), the first thing we want to do is find our derivative using the product rule. Remember, the product rule tells us that we have our left function and we have our right function. Recalling our memory tool, we know that "left \( d \) right plus right \( d \) left" is our derivative formula. Therefore, if we consider these functions, we have the left function, \( f(x) \), multiplied by the derivative of the right function, \( g'(x) \), then added to the right function, \( g(x) \), times the derivative of the left function, \( f'(x) \). Now, this is our derivative.

Let's see what happens when we plug 2 in because our goal here is to find \( h'(2) \). Plugging 2 in, we take \( f(2) \) multiplied by \( g'(2) \) and add \( g(2) \) multiplied by \( f'(2) \). Using the information provided in the problem, we find that \( f(2) = 4 \), \( g'(2) = 0 \), \( g(2) = 3 \), and \( f'(2) = -1 \). Calculating, \( 4 \times 0 + 3 \times (-1) = 0 - 3 \) which simplifies to \( h'(2) = -3 \) for this scenario.

Moving onto part B, we find \( h'(2) \) when \( h(x) = \frac{f(x)}{g(x)} \) using the quotient rule. We have our high function \( f(x) \) on top and our low function \( g(x) \) on the bottom. Following the formula for the quotient rule, "low d high minus high dee low over the square of what's below," \( h'(x) \) is equal to \( g(x) \times f'(x) - f(x) \times g'(x) \) all divided by \( g(x)^2 \). Let's substitute \( x = 2 \) into our expression. With \( f(2) = 4 \), \( g'(2) = 0 \), \( g(2) = 3 \), and \( f'(2) = -1 \), we calculate \( 3 \times (-1) - 4 \times 0 \) all over \( 3^2 \) which simplifies to \( -3/9 = -1/3 \).

This is our answer for \( h'(2) \) under this scenario and it serves as our final answer for part B of the problem. Such a problem may appear intimidating, but with a systematic approach, it solves quite smoothly. If you have any questions, feel free to ask. I'll see you in the next problem.

The Quotient Rule Example 5

In this problem, we're told that the population of a species of bird is given by the function \( p(t) = \frac{6t+5}{0.3t^2+2} \), where \( t \) is specifically time measured in years. Now we are asked here to find \( p'(t) \), the derivative, and also interpret its meaning. So we are using our math skills and our derivative skills here and combining that with some critical thinking in order to determine what this math we're doing means in the context of this problem. So let's go ahead and get started in first finding our derivative here, \( p'(t) \). Now looking at my function here, since I have one function on the top \( 6t+5 \) divided by another function on the bottom \( 0.3t^2+2 \), I need to use the quotient rule.

Remember, the quotient rule tells us \( \text{low} \cdot \text{d high} - \text{high} \cdot \text{d low} \) over the square of what's below is our derivative. So let's go ahead and apply that quotient rule here. We're starting with that low function, which is \( 0.3t^2 + 2 \) times the derivative of that high function. The derivative of \( 6t+5 \) is just 6. And then here, we're subtracting high \( d \) low.

That is our high function, \( 6t+5 \) times the derivative of that bottom function \( 0.3t^2+2 \). The derivative of this function is just going to be \( 0.6t \) using the power rule. Now that's our full numerator. Now for our denominator, it's over the square of what's below, so I'm taking that function on the bottom, \( 0.3t^2 + 2 \), and I'm going to fully square that because remember it's the square of what's below, not just what's below. So from here, we can do some simplification.

We want to go ahead and distribute this 6 in here to see if we can end up canceling anything out and then also this \( 0.6t \). So in the top, I have \( 6 \times 0.3 \). That's going to give me \( 1.8t^2 + 12 \) for that first term and then I am subtracting this \( 0.6t \) distributed throughout. \( 6 \times 0.6 \) is going to give me \( 3.6t^2 \) because I have a \( t \) in both of those terms and then a plus \( 5 \times 0.6t \) which is going to give me \( 3t \). Now from here I want to distribute this negative in here just to go ahead and do that all in one step.

So just know that both of these terms are going to be negative because we have distributed that. Then my denominator stays the same at \( (0.3t^2+2)^2 \). And I can go ahead and do some further simplification here. I have some like terms. I have this \( 1.8t^2 \).

I have a minus \( 3.6t^2 \). So this becomes a negative \( 1.8t^2 \). And then a minus \( 3t \), just writing this in standard form of a polynomial, plus \( 12 \). Then my denominator stays the same at \( (0.3t^2+2)^2 \). All of that is squared and this is my final derivative here \( p'(t) \).

Now we have answered the first part of our question which is to find \( p'(t) \), our derivative, but we want to answer that second part too. We want to interpret what this actually means. Now this is a population function. We're told that this is a function that represents the population of a species of bird where \( t \) is measured in years. So in finding the derivative here, remember that the derivative is a rate of change, so we found the rate of change of the population of this bird.

So if we were to plug in a value and find, say, \( p'(2) \), we would be finding the rate of change of this bird population at the time of 2 years. Now if you have any questions here, feel free to let us know and I will see you in the next video.