You may have seen how to take the derivative of constants, powers of \(x\), and polynomials, but how would you take the derivative of something like this?
In this video, I’m going to show you how to take the derivative when a given function includes the multiplication or division of two or more differentiable functions.
First, let’s talk about how to handle derivatives involving the multiplication of functions. To take such derivatives, we’ll need something called the product rule.
The Product Rule
The product rule states that if you want to take the derivative of two functions multiplied together, start by writing the first function and multiplying it by the derivative of the second function. Then, add the second function multiplied by the derivative of the first function.
Mathematicians write the product rule in this generalized form:
In this case, \(f\) and \(g\) are two things being multiplied together, and \(f’\) and \(g’\) are their respective derivatives.
Product Rule Examples
Example 1
Let’s try some examples of using the product rule. Let’s take a look at the function:
To find the derivative of this function, remember that the product rule says that the derivative of (fg)=fg’+gf’[/latex]. In this case, \(f\) is equal to the first function, \(x^{4}\), and \(g\) is equal to the second function, which is \(ln(x)\).
We need to write down \(fg’\). In this case, that will be \(x^\times \frac{1}{x}\), since the derivative of \(ln(x)\) is \(\frac{1}{x}\).
Together, these can just be reduced to \(x^{3}\).
The last step in using the product rule is writing the term \(gf’\). Remember, in this problem \(g=ln(x)\), so we will write that multiplied by the derivative of \(f\). 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
The derivative of the function \(p(x)=x^{4} ln(x)\) is then equal to \(gf’=x^{3}+ln(x)\cdot 4x^{3}\).
Example 2
Let’s try another example. Use the product rule to find the derivative of \(q(x)=e^xcos(x)\). Remember, the product rule says that the derivative will equal the first function times the derivative of the second, plus the second function times the derivative of the first.
Let’s write down the first function, \(e^{x}\), multiplied by the derivative of the second function. Since the derivative of \(cos(x)\) is \(-sin(x)\), we get:
Then, we are going to add the second function multiplied by the derivative of the first function. Since \(e^{x}\) is its own derivative, \(gf’=cos(x)\cdot e^x\).
Combining both parts, we get:
We can clean this up a little bit by factoring the \(e^{x}\) out from both terms and rearranging a bit:
Example 3
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, we can combine 2 and \(x\) as \(f\) and leave \(e^x\) as \(g\). Using these definitions, the derivative of \(r(x)\) is equal to the first piece times the derivative of the second, which is \(2x\cdot e^{x}\), plus the second piece times the derivative of the first, which is \(e^x\cdot 2\).
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.
The Quotient Rules
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 that 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.
Remembering the Quotient Rule
Let’s say we want to take the derivative of this fraction: \(\frac{\text{High}}{\text{Low}}\).
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:
This little saying is a memory tool to help you remember the quotient rule.
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 equal to the starting denominator squared.
To use the letters \(f\) and \(g\), the quotient rule can be written in this way:
Remember: “Low d High, minus High d Low, over the square of what’s below.”
Quotient Rule Examples
Example 1
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 \(s'(x)=(x^{3}-1)\cdot 9\).
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.
We can distribute our 9 into our parentheses and get:
Then, we can combine our \(x^{3}\) terms and we’ll get:
Now, the bottom of the derivative equals “the square of what’s below,” so that’ll be \((x^{3}-1)^{2}\).
The derivative of \(s(x)\) is then:
Example 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 \(sin(x)\) times the derivative of \(e^x\), which is still \(e^x\), minus \(e^{x}\), times the derivative of \(sin(x)\), which is \(cos(x)\).
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.
We can factor the numerator to clean it up a little, so that we have:
And that’s our derivative!
Example 3
Let’s try one last example. This one will require use of both the product rule and the quotient rule.
Find the derivative of:
Notice that this function is in fraction form, with a product in the numerator. Let’s label the top \(f\) and the bottom \(g\) to make it easier to keep track of everything.
\(g=7x^{2}+2\)
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 \(sin(x)\), which is \(cos(x)\), and then we’ll add \(sin(x)\) times the derivative of \(x\), which is 1.
Since \(g\) is a polynomial, \(g’=14x\):
Now that we have \(f’\) and \(g’\), we can use the quotient rule. “Low d High” means \(gf’\), so we start by writing:
Then, subtract “High d Low” or \(fg’\).
Then the denominator of the derivative will be \(g^{2}\), which is \((7x^2+2)^2\)
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!
