Answer
$f'(4)=\frac{-3}{16}$
Work Step by Step
$f(x)=\frac{3}{x}$
$f'(c)=\lim\limits_{x \to c}\frac{f(x)-f(c)}{x-c}$
$f'(4)=\lim\limits_{x \to 4}\frac{\frac{3}{x}-\frac{3}{4}}{x-4}$
$f'(4)=\lim\limits_{x \to 4}\frac{\frac{12-3x}{4x}}{x-4}$
$f'(4)=\lim\limits_{x \to 4}\frac{\frac{-3(x-4)}{4x}}{x-4}$
$f'(4)=\lim\limits_{x \to 4}\frac{-3}{4x}$
$f'(4)=\frac{-3}{4(4)}$
$f'(4)=-\frac{3}{16}$
