Answer
$\frac{-1}{49}$
Work Step by Step
$f(x)=\frac{1}{x+4}$
$c=3$
$f'(c)=\lim\limits_{x \to c}\frac{f(x)-f(c)}{x-c}$
$f'(3)=\lim\limits_{x \to 3}\frac{\frac{1}{x+4}-\frac{1}{3+4}}{x-3}=\lim\limits_{x \to 3}\frac{\frac{1}{x+4}-\frac{1}{7}}{x-3}=\lim\limits_{x \to 3}\frac{(\frac{7-(x+4)}{7(x+4)})}{x-3}=\lim\limits_{x \to 3}{\frac{-(x-3)}{7(x+4)(x-3)}}=\lim\limits_{x \to 3}{\frac{-(x-3)}{7(x+4)(x-3)}}=\lim\limits_{x \to 3}\frac{-1}{7x+28}=\frac{-1}{7(3)+28}=\frac{-1}{21+28}=\frac{-1}{49}$
