# Product and Quotient Rule

Up to this point, you’ve learned how to take the derivative of constants, powers of $$x$$, and polynomials, but how would you take the derivative of something like this?

$$f(x)=\frac{6x}{2x^{2}+5}+x\text{ }sin(x)$$

In this video, I’m going to show you how to take derivatives where multiplication, division, or both are present!

Let’s first talk about how to handle derivatives involving multiplication. To take such derivatives, we’ll need something called the product rule.

The product rule states that if you want to take the derivative of two factors multiplied together, start by writing the first factor and multiplying it with the derivative of the second factor. Then, add the second factor multiplied with the derivative of the first factor. Mathematicians usually write the product rule in this generalized form:

$$\frac{d}{dx}f\text{ }g=f\text{ }g’+gf’$$

where $$f$$ and $$g$$ are two things being multiplied together, and $$f’$$ and $$g’$$ are their respective derivatives.

Let’s try some examples of using the product rule. Let’s take a look at the function:

$$p(x)=x^{4} ln(x)$$

To find the derivative of this function, remember that the product rule says that the derivative of $$f$$ times $$g$$ is $$f$$ times $$g’$$, plus $$g$$ times $$f’$$. In this case, $$f$$ is equal to the first factor in the function, $$x^{4}$$, and $$g$$ is equal to the second factor, which is $$ln(x)$$.

We need to write down $$f$$ times $$g’$$, the first factor times the derivative of the second. In this case, that will be $$x^{4}$$ times $$\frac{1}{x}$$, since the derivative of $$ln(x)$$ is $$\frac{1}{x}$$.

$$p'(x)=x^{4}\cdot \frac{1}{x}$$

Together, of course, these can just be reduced to $$x^{3}$$.

$$p'(x)=x^{3}…$$

Then, write a plus sign since the product rule tells us to add $$f$$ times $$g’$$ with $$g$$ times $$f’$$:

$$p'(x)=x^{3}+…$$

The last step in using the product rule is writing the term $$g$$ times $$f’$$. Remember, in this problem $$g=ln(x)$$, so we will write that multiplied by the derivative of the first factor. Since the derivative of $$x^{4}$$ is $$4x^{3}$$, we write $$ln(x)\cdot 4x^{3}$$. The derivative of the function $$p(x)=x^{4}ln(x)$$ is then equal to

$$p'(x)=x^{3}+ln(x)\cdot 4x^{3}$$

Let’s try another example. Use the product rule to find the derivative of qx=excosx. Remember, the product rule says that the derivative will equal the first factor times the derivative of the second, plus the second factor times the derivative of the first.

Let’s write down the first factor, $$e^{x}$$, multiplied by the derivative of the second factor. Since the derivative of $$cos(x) is -sin(x)$$, we get:

$$q'(x)=e^{x}[-sin(x)]…$$

Then we are going to add the second factor multiplied by the derivative of the first factor. Since $$e^{x}$$ is its own derivative, we get $$cos(x)$$ times $$e^{x}$$:

$$q'(x)=e^{x}-sin(x))+cos(x)e^{x}$$

We can clean this up a little bit by factoring the $$e^{x}$$ out from both terms and writing:

$$q'(x)=e^{x}[-sin(x)+cos(x)]$$

Let’s take a look at one more example. Use the product rule to find the derivative of $$r(x)=2xe^{x}$$. Now this is technically three things multiplied together: 2, $$x$$, and $$e^{x}$$. But of course, we know how to take the derivative of 2 times $$x$$ already, so we can treat those as one piece together. In this example, $$f$$ is $$2x$$ and $$g$$ is $$e^{x}$$. The derivative of $$r(x)$$ should then be equal to the first piece times the derivative of the second:

$$r'(x)=2x\cdot e^{x}…$$

Plus the second piece times the derivative of the first:

$$r'(x)=2x\cdot e^{x}+e^{x}\cdot 2$$

Simplifying, we get $$2xe^{x}+2e^{x}$$, or we can factor out the $$2e^{x}$$ and get $$2e^{x}(x+1)$$.

Some students have difficulty committing the product rule to memory. However, with some practice on your own, you should start to become comfortable with using it.

While the product rule helps us find derivatives of functions containing multiplication, we have another tool called the quotient rule to help us find derivatives of functions containing division.

The quotient rule states the derivative of a fraction is a new fraction, whose numerator equals the original fraction’s bottom times its top’s derivative, minus the top times the bottom’s derivative, and whose denominator equals the original fraction’s bottom squared. If you didn’t catch all that, don’t worry. I have a way to help you remember the quotient rule.

Let’s say we want to take the derivative of this fraction:

$$\frac{High}{Low}$$

By “high” I mean to indicate the numerator, and by “low,” the denominator. To get the derivative of this fraction, the quotient rule can be remembered in this way:

$$\frac{\text{“Low d High, minus High d Low,}} {\text{over the square of what’s below.”}}$$

This little saying is a memory tool to help you remember the quotient rule. Let me say it again so you can become familiar with it:

$$\frac{\text{“Low d High, minus High d Low,}} {\text{over the square of what’s below.”}}$$

Now here’s what that means. The derivative of a fraction will be another fraction. The top of this fraction is given by the first part of the rhyme: “Low d High, minus High d Low.” Write the denominator—“Low”—and multiply by the derivative of the numerator—that’s what I mean by “d High”. Then subtract the numerator—“High”—multiplied with the derivative of the denominator—“d Low.”

The second line of the rhyme says, “over the square of what’s below.” This means that the denominator of the solution is simply equal to the starting denominator, squared.

To use the letters $$f$$ and $$g$$, the quotient rule can be written in this way:

$$\frac{d}{dx}\frac{f}{g}=\frac{gf’-fg’}{g^{2}}$$

Remember:

$$\frac{\text{“Low d High, minus High d Low,}} {\text{over the square of what’s below.”}}$$

Let’s try this out on an example. Use the quotient rule to find the derivative of the function

$$s(x)=\frac{9x}{x^{3}-1}$$

Since this function is in fraction form, we know that its derivative will also be a fraction. The top of this fraction will equal “low d high minus high d low.” In this case, “Low” equals $$x^{3}-1$$, and “d High” equals the derivative of $$9x$$, which is 9. So the first thing we have is $$s'(x)=(x^{3}-1)\cdot 9$$.

Then we are going to subtract “high d low,” which is $$9x$$ times the derivative of the bottom, which is $$3x^{2}$$. So the numerator of the derivative will be equal to $$s'(x)=(x^{3}-1)\cdot 9-9x(3x^{2})$$. Let’s multiply and collect like terms to clean this up a little bit.

$$s'(x)=(x^{3}-1)\cdot 9-9x(3x^{2})$$

So we can distribute our 9 into our parentheses and get:

$$s'(x)=9x^{3}-9-27x^{3}$$

Then we can combine our $$x^{3}$$ terms and we’ll get:

$$s'(x)=-18x^{3}-9$$

Now, the bottom of the derivative equals “the square of what’s below,” so that’ll be $$(x^{3}-1)^{2}$$. The derivative of $$\frac{9x}{x^{3}-1}$$ is then:

$$s'(x)=\frac{-18x^{3}-9}{(x^{3}-1)^{2}}$$

Let’s try another example with the quotient rule. Find the derivative of $$t(x)=\frac{e^{x}}{sin(x)}$$. To find $$t'(x)$$, we know we are going to have a fraction. The top will equal “low d high minus high d low,” so that’s “low” $$sin(x)$$ times the derivative of $$e(x)$$, which is still $$e(x)$$:

$$t'(x)=sin(x)\cdot e^{x}…$$

Minus “high”, $$e^{x}$$, times the derivative of $$sin(x)$$, which is $$cos(x)$$.

$$t'(x)=sin(x)\cdot e^{x}-e^{x}\cdot cos(x)…$$

Then the bottom of the fraction will equal “low squared,” so $$sin(x)$$ squared. Now remember, you can write $$sin(x)^{2}$$ as $$sin^{2}(x)$$, this is just a different way of writing the same thing.

$$t'(x)=\frac{sin(x)\cdot e^{x}-e^{x}\cdot cos(x)}{sin^{2}(x)}$$

We can factor the numerator to clean it up a little, so that we have:

$$t'(x)=\frac{e^{x}sin(x)-cos(x))}{sin^{2}(x)}$$

And that’s our derivative!

Let’s try one last example. This one will require use of both the product rule and the quotient rule. Find the derivative of:

$$u(x)=\frac{x\text{ }sin(x)}{7x^{2}+2}$$

We can clearly see that this function is in fraction form, with a product in its numerator. Let’s label the top $$f$$ and the bottom $$g$$ to keep track of everything easier.

$$f=x\text{ }sin(x)$$

$$g=7x^{2}+2$$

Now, the quotient rule tells us that the derivative’s numerator equals “low d high minus high d low,” so let’s go ahead and figure out what $$f’$$ and $$g’$$ are equal to now.

To get $$f’$$, we need to use the product rule. First, we write $$x$$ times the derivative of sine, and then we’ll add sine times the derivative of $$x$$.

$$f’=x\text{ }cos{x}+sin{x}$$

The derivative of $$g$$ is pretty simple since it is just a polynomial.

$$g’=14x$$

Now that we have $$f’$$ and $$g’$$, we can use the quotient rule. “Low d High” means $$g$$ times $$f’$$, so we start by writing:

$$u'(x)=(7x^{2}+2)(xcos(x)+sin(x))…$$

Then, subtract “High d Low” or $$f$$ times $$g’$$.

$$u'(x)=(7x^{2}+2)(xcos(x)+sin(x))-(xsin(x))(14x)…$$

Then the bottom of the derivative will simply equal $$g^{2}$$.

$$u’x=\frac{(7x^{2}+2)(xcos(x)+sin(x))-(xsin(x))(14x)}{(7x^{2}+2)^{2}}$$

This can be slightly simplified by multiplying a few of the terms.

$$u'(x)=\frac{(7x+^{2})(xcos(x)+sin(x))-14x^{2}sin(x)}{(7x^{2}+2)^{2}}$$

The product rule and quotient rule are fairly easy to work with once they are committed to memory, and the easiest way to memorize them is to work through some examples on your own.

Thanks for watching, and happy studying!

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by Mometrix Test Preparation | This Page Last Updated: January 25, 2023