## Calculus 10th Edition

$f'(x)=\frac{8}{(x+4)^2};f'(3)=\frac{8}{49}$
Using the quotient rule: $f’(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=(x-4); u'(x)=1$ $v(x)=(x+4); v'(x)=1$ $f'(x)=\frac{(1)(x+4)-(x-4)(1)}{(x+4)^2}=\frac{8}{(x+4)^2}$. $f'(3)=\frac{8}{(3+4)^2}=\frac{8}{49}$ Alternative way to find the derivative: The following is a trick that you will use to integrate some complicated functions later on. Make the numerator the same as the denominator by adding and subtracting the same number from the numerator as follows: $f(x)=\frac{x-4}{x+4}=\frac{x-4+8-8}{x+4}=\frac{(x+4)-8}{x+4}=\frac{x+4}{x+4}-\frac{8}{x+4}=1-\frac{8}{x+4}$ Now $f'(x)$ can be found using the Power Rule: $f'(x)=(1)'-(8((x+4)^{-1})')=\frac{8}{(x+4)^2}$