Product Rule Derivatives: Easy, Medium, and Hard
Simple overview of examples that use many derivatives rules (with an emphasis on the product rule).
Derivative Product Rule
We use the product rule when a function is written as one function multiplied by another function.
\[\frac{d}{dx}\left[f(x)g(x)\right]=f(x)g'(x)+g(x)f'(x)\]A good way to remember the structure is
\[\text{first}\cdot \text{derivative of second}+\text{second}\cdot \text{derivative of first}\]Product Rule Only
Consider
\[f(x)=x\tan(x)\]Think of this as the product of two functions
\[f(x)=\underbrace{x}_{\text{first}}\underbrace{\tan(x)}_{\text{second}}\]Their derivatives ar
\[\frac{d}{dx}(x)=1\]and
\[\frac{d}{dx}\left(\tan(x)\right)=\sec^2(x)\]Now plug into the product rule (\(\frac{d}{dx}\left[f(x)g(x)\right]=f(x)g'(x)+g(x)f'(x)\))
\[f'(x)=x\sec^2(x)+\tan(x)\cdot 1\]So
\[f'(x)=x\sec^2(x)+\tan(x)\]Watch this part: easy example at 0:49.
Medium Examples: Product Rule Plus Another Rule
Each of the medium level examples uses the product rule paired with another rule.
Consider the following
\[f(x)=x^3e^x\]The two functions multiplied together are $x^3$ and $e^x$. Their derivatives are
\[\frac{d}{dx}(x^3)=3x^2 \quad \text{ this step uses the power rule}\]and
\[\frac{d}{dx}(e^x)=e^x\]Apply the product rule
\[f'(x)=x^3e^x+e^x(3x^2)\]You can leave the answer like that, or factor out $e^x$
\[f'(x)=e^x(x^3+3x^2)\]Ultimately, the structure of this derivative comes from the product rule but within the process of finding the derivative the power rule was also used.
Watch this part: product rule with the power rule at 2:33.
Another medium-level example in the video is:
\[f(x)=\sin(x)e^{\cos(x)}\]This still uses the product rule first, but the derivative of $e^{\cos(x)}$ needs the chain rule
\[\frac{d}{dx}\left(e^{\cos(x)}\right)=e^{\cos(x)}(-\sin(x))\]Then
\[f'(x)=\sin(x)\left[-\sin(x)e^{\cos(x)}\right]+e^{\cos(x)}\cos(x)\]A cleaner factored form is
\[f'(x)=e^{\cos(x)}\left(\cos(x)-\sin^2(x)\right)\]So this example used the product rule and when taking the derivative of the second function we had to use the chain rule.
Watch this part: product rule with the chain rule at 5:21.
Harder Examples: Decide The Outer Rule First
What happens when we have products within fractions (quotients)?
Consider
\[f(x)=\frac{x\sin(x)}{\ln(x)}\]The outer structure is a quotient, so start with the quotient rule. Let
\[N=x\sin(x)\]and
\[D=\ln(x)\]The numerator $N$ is itself a product, so
\[N'=x\cos(x)+\sin(x) \quad \text{(here is where we use the product rule)}\]The denominator derivative is
\[D'=\frac{1}{x}\]Now use the quotient rule
\[f'(x)=\frac{N'D-ND'}{D^2}\]So plugging everything we found into the quotient rule guide we get
\[f'(x)=\frac{\left(x\cos(x)+\sin(x)\right)\ln(x)-x\sin(x)\left(\frac{1}{x}\right)}{\left(\ln(x)\right)^2}\]Watch this part: quotient rule with product rule inside at 8:44.
The Big Final Pattern
A later example combines the product rule, quotient rule, chain rule, and power rule:
\[f(x)=\frac{e^{\sin(x)}(x^8+x)}{\tan(x)}\]Because the whole expression is a quotient, start by naming the top ($N$ for numerator) and bottom ($D$ for denominator) functions
\[N=e^{\sin(x)}(x^8+x)\]and
\[D=\tan(x)\]The numerator is a product, so find $N’$ with the product rule:
\[N'=e^{\sin(x)}\cos(x)(x^8+x)+e^{\sin(x)}(8x^7+1) \quad \text{(in this derivative we use the product, chain, and power rules)}\]The denominator derivative is
\[D'=\sec^2(x)\]Now place those pieces into the quotient rule
\[f'(x)=\frac{\left[e^{\sin(x)}\cos(x)(x^8+x)+e^{\sin(x)}(8x^7+1)\right]\tan(x)-e^{\sin(x)}(x^8+x)\sec^2(x)}{\tan^2(x)}\]Watch this part: hard example at 26:44.
Timestamp Guide
More Practice
For the full list of timestamped practice problems from this video, use the existing archives: