3.2: Product and Quotient Rule

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Throughout this module, if something does not exist, write DNE in the answer box.

Recap Video

Take a look at the following video which recaps the ideas from the section.

Example Video

Below are two videos showing worked examples.

First example:

Second example:

Problems

Here are the derivative rules you should know from the section:

Let $f(x)$ and $g(x)$ be differentiable functions. Then:

(Product Rule) $[f(x)g(x)]' = f'(x)g(x)+f(x)g'(x)$ .
(Quotient Rule) $\displaystyle \left (\frac {f(x)}{g(x)}\right )' = \frac {g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}$ .

If $f(x) = x^2e^x$ , find $f'(x)$ .

Notice that $f(x)$ is the product of two functions, let’s call them $g(x)$ and $h(x)$ (so $f(x) = g(x)h(x)$ ) with $g(x) = x^2,\quad h(x) = \answer {e^x}.$ The product rule says $f'(x) = g'(x)h(x)+g(x)h'(x).$ Notice $g'(x) = \answer {2x}$ by the power rule, and $h'(x) = \answer {e^x}$ . Therefore, $f'(x) = \answer {2xe^x+x^2e^x}.$

If $\displaystyle f(x) = \frac {e^x+x}{x^2+1}$ , evaluate $f'(x)$ .

This function is a quotient of two functions $g(x)$ and $h(x)$ . We can write $f(x) =g(x)/h(x)$ where $g(x) = e^x+x,\quad h(x) = \answer {x^2+1}.$ The quotient rule says $f'(x) = \frac {h(x)g'(x)-g(x)h'(x)}{h(x)^2}.$ Notice $g'(x) = \answer {e^x+1},\quad h'(x) = \answer {2x}.$ Plugging everything in, we get $f'(x) = \answer {\frac {(x^2+1)(e^x+1)-2x(e^x+x)}{(x^2+1)^2}}.$

If $F(s) = (s+s^2)(2s-e^s)$ , then $F'(s) = \answer {(1+2s)(2s-e^s)+(s+s^2)(2-e^s)}.$

Use the product rule.

If $f(t) = \frac {e^t}{e^t-t}$ , then $f'(t) = \answer {\frac {(e^t-t)e^t - e^t(e^t-1)}{(e^t-t)^2}}.$

Suppose $f(x) = x^3g(x)$ with $g(1) = 1$ and $g'(1) = 3$ . Then $f'(1) = \answer {6}$ .

We can use the product rule to differentiate $f(x)$ . Doing so gives $f'(x) = 3x^2g(x)+x^3g'(x).$ Now plug in $x=1$ .

Consider $f(x) = \frac {x}{x^2+1}$ . How many points are there on the graph of $y = f(x)$ where the tangent line is horizontal? $\answer {2}$ .

Use the quotient rule to differentiate and simplify the derivative.

The only way for a fraction to equal $0$ is if the numerator equals $0$ .

List them in increasing order: $x = \answer {-1} \quad \text { and }\quad x = \answer {1}.$