# 3.2 Product, Quotient Rules

## Product, Quotient Rules

The basic rules will let us tackle simple functions. But what happens if we need the derivative of more complicated functions such as the product or quotients of functions?

### Example 1

Find the derivative of [latex]h(x)=\left(4x^3-11\right)(x+3)[/latex]

This function is not a simple sum or difference of polynomials. It’s a product of polynomials. We can simply multiply it out to find its derivative:

\[ \begin{align*}

h(x)=& \left(4x^3-11\right)(x+3)\\

=& 4x^4-11x+12x^3-33\\

h'(x)=& 16x^3-11+36x^2

\end{align*} \]

Now suppose we wanted to find the derivative of \[f(x)=\left(4x^5+x^3-1.5x^2-11\right)\left(x^7-7.25x^5+120x+3\right)\]

This function is not a simple sum or difference of polynomials. It’s a product of polynomials. We could ‘simply’ multiply it out to find its derivative as before – who wants to volunteer? Nobody?

We’ll need a rule for finding the derivative of a product so we don’t have to multiply everything out.

It would be great if we can just take the derivatives of the factors and multiply them, but unfortunately that won’t give the right answer. To see that, consider finding derivative of [latex]h(x)=\left(4x^3-11\right)(x+3)[/latex]. We already worked out the derivative, it is [latex]h'(x)=16x^3-11+36x^2[/latex]. What if we try differentiating the factors and multiplying them? We’d get [latex]h'(x)=\left(12x^2\right)(1)=12x^2[/latex], which is radically different from the correct answer.

The rules for finding derivatives of products and quotients are a little complicated, but they save us the much more complicated algebra we might face if we were to try to multiply things out. They also let us deal with products where the factors are not polynomials. We can use these rules, together with the basic rules, to find derivatives of many complicated looking functions.

### Derivative Rules: Product and Quotient Rules

In what follows, [latex]f[/latex] and [latex]g[/latex] are differentiable functions of [latex]x[/latex].

#### Product Rule

$$\frac{d}{dx}\left( f\cdot g \right)=f’\cdot g+f\cdot g’$$

The derivative of the first factor times the second left alone, plus the first left alone times the derivative of the second.

The product rule can extend to a product of several functions; the pattern continues – take the derivative of each factor in turn, multiplied by all the other factors left alone, and add them up: \[\frac{d}{dx}\left( f\cdot g\cdot h \right)=f’\cdot g\cdot h+f\cdot g’\cdot h+f\cdot g\cdot h’\]

#### Quotient Rule

\[\frac{d}{dx}\left( \frac{f}{g} \right)=\frac{f’\cdot g-f\cdot g’}{g^2}\]

The numerator of the result resembles the product rule, but there is a minus instead of a plus; the minus sign goes with the [latex]g’[/latex]. The denominator is simply the square of the original denominator – no derivatives there.

### Example 2

Find the derivative of [latex]h(x)=\left(4x^3-11\right)(x+3)[/latex]

This is the same function we found the derivative of in Example 1, but let’s use the product rule and check to see if we get the same answer. For this first example, we will provide a lot more detail and steps than one usually actually shows when working a problem like this.

Notice we can think of [latex]h(x)[/latex] as the product of two functions [latex]f(x)=4x^3-11[/latex] and [latex]g(x)=x+3[/latex]. Finding the derivative of each of these, \[ f'(x)=12x^2 \ \text{and}\ g'(x)=1. \]

Using the product rule,

\[ \begin{align*}

h'(x)=& (f’)(g)+(f)(g’) \\

=& \left(12x^2\right)(x+3)+\left(4x^3-11\right)(1)

\end{align*} \]

To check if this is equivalent to the answer we found in Example 1 we could simplify:

\[ \begin{align*}

h'(x)=& \left(12x^2\right)(x+3)+\left(4x^3-11\right)(1) \\

=& 12x^3+36x^2+4x^3-11 \\

=& 16x^3+36x^2-11

\end{align*} \]

From this, we can see the answers are equivalent.

### Example 3

Find the derivative of [latex]F(t)=e^t\ln(t)[/latex]

This is a product, so we need to use the product rule. I like to put down empty parentheses to remind myself of the pattern; that way I don’t forget anything: \[F'(t)=(\ )(\ )+(\ )(\ )\]

Then I fill in the parentheses – the first set gets the derivative of , the second gets left alone, the third gets left alone, and the fourth gets the derivative of \[F'(t)=\left(e^t\right)\left(\ln(t)\right)+\left(e^t\right)\left(\frac{1}{t}\right)=e^t\ln(t)+\frac{e^t}{t}.\]

Notice that this was one we could *not* have done by multiplying out.

### Example 4

Find the derivative of [latex]y=\frac{x^4+4^x}{3+16x^3}[/latex].

This is a quotient, so we need to use the quotient rule. Again, you find it helpful to put down the empty parentheses as a template: \[y’=\frac{(\ )(\ )-(\ )(\ )}{(\ )^2}\]

Then fill in all the pieces: \[y’=\frac{\left(4x^3+\ln(4)\cdot 4^x \right)\left(3+16x^3 \right)-\left(x^4+4^x \right)\left(48x^2 \right)}{\left(3+16x^3 \right)^2}\]