Product/quotient rule [3.3]

Area of a rectangle

Let's say that for a particular rectangle, $$w(t)=t;\ \ \ h(t)=t^2$$ Then the area of the rectangle will be a function of time, given by: $$A(t)=w(t)h(t)=t*t^2=t^3$$

So $$A'(t)=3t^2$$

Make a guess...Is $A'(t) maybe the same as $w'(t)h't$?? $$w'(t)h'(t)=1*2t=2t,$$ ...so, no.

Can we find the derivative of the product of two functions some other way?...

Proof of product rule

Given $A(t)=f(t)g(t)$, what is $A'(t)$? $$\begineq A'(t)=&\lim_{h\to 0} \frac{A(t+h)-A(t)}{h} =\lim_{h\to 0} \frac{f(t+h)g(t+h)-f(t)g(t)}{h}\\ =&\lim_{h\to 0} \frac{f(t+h)g(t+h)\color{red}+f(t)g(t+h) \color{blue}-f(t)g(t+h) \color{black}-f(t)g(t)}{h} \\ =&\lim_{h\to 0} \frac{f(t+h)g(t+h) \color{blue}-f(t)g(t+h) \color{red}+f(t)g(t+h) -\color{black}f(t)g(t)}{h} \\ =&\lim_{h\to 0}\left[ \frac{f(t+h)-f(t)}{h}g(t+h) +f(t)\frac{g(t+h)-g(t)}{h}\right] \\ =&\lim_{h\to 0} \frac{f(t+h)-f(t)}{h}\cdot \lim_{h\to 0} g(t+h) + \lim_{h\to 0}f(t) \cdot \lim_{h\to 0}\frac{g(t+h)-g(t)]}{h} \\ =&f'(t)g(t)+f(t)g'(t). \endeq $$

Product rule: $$\frac{d}{dx}[f(x)\cdot g(x)]=f'(x)g(x)+f(x)g'(x).$$

How does it do with the rectangle where $w(t)=t$ and $h(t)=t^2$? $$\begineq A'(t)=&w'(t)h(t)+w(t)h'(t)\\ =&1\cdot t^2+t\cdot 2t\\ =&\mathbf{3t^2}. \endeq$$

Example: $A(x)=x^4\ln(x+3)$.

Here, $f(x)=x^4 \Rightarrow f'(x)=4x^3$,
and $g(x)=\ln(x+3) \Rightarrow g'(x)=\frac{1}{x+3}(x+3)'=\frac{1}{x+3}$, so...

$$A'(x)=4x^3\ln(x+3)+\frac{x^4}{x+3}.$$

Quotient rule

If $q(x)=\frac{f(x)}{g(x)}$, what is $q'(x)$?

Write this as $q(x)=f(x)h(x)$ where $h(x)=1/g(x)=[g(x)]^{-1}$. Then the derivative of $h$ can be found with the chain rule: $$h'(x)=-1*[g(x)]^{-2}\cdot g'(x) = -\frac{g'(x)}{(g(x))^2}.$$

So $$\begineq q'(x)=& f'(x)h(x)+f(x)h'(x)=\frac{f'(x)}{g(x)} -f(x)\frac{g'(x)}{(g(x))^2}\\ =&\frac{f'(x)g(x)}{(g(x))^2} -f(x)\frac{g'(x)}{(g(x))^2}\\ =&\frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}. \endeq $$

Quotient rule: $$\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}$$

Example: $q(x)=\frac{7x^3}{e^x+2}$

Steps:

1. Write $q(x)=\frac{f(x)}{g(x)}$ and identify the numerator and denominator functions: Apparently $$f(x)=7x^3 \text{ and } g(x)=e^x+2.$$
2. Calculate the derivatives of $f$ and $g$: $$f'(x)=7*(3x^2)=21x^2.$$ $$g'(x)=e^x+0=e^x.$$
3. Use the quotient rule... $$\begineq q'(x)=&\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}\\ =&\frac{21x^2(e^x+2)-7x^3e^x}{[e^x+2]^2}\\ =&\frac{7x^2(3e^x+6-xe^x)}{e^{2x}+4e^x+4}\\ &\text{or}&\\ =&\frac{7x^2(e^x(3-x)+6)}{[e^x+2]^2}\\ \endeq$$ Checking on WolframAlpha

   Checking graphically...
   
   

   More examples (@ UCDavis, with solutions).