Answer
$$f'(x)=-\frac{1}{(3-x)^2}$$
$$f''(x)=\frac{2}{(3-x)^3}$$
Work Step by Step
$$f(x)=\frac{1}{3-x}=(3-x)^{-1}$$
Differentiating with respect to $x$, using power rule,
$$f'(x)=-1(3-x)^{-2}=-\frac{1}{(3-x)^2}$$
Differentiating again with respect to $x$, using power rule,
$$f''(x)=(-1)(-2)(3-x)^{-3}=\frac{2}{(3-x)^3}$$.
