Answer
$f^{n}(x)=\frac{n!}{(2-x)^{n+1}}$
Work Step by Step
$$f'(x)=\frac{1'(2-x)-1(2-x)'}{(2-x)^{2}}=\frac{1}{(2-x)^{2}}=\frac{1!}{(2-x)^{1+1}}$$
$$f^{2}(x)=\frac{1'(2-x)^{2}-1((2-x)^{2})'}{(2-x)^{4}}=\frac{2}{(2-x)^{3}}=\frac{2!}{(2-x)^{2+1}}$$
$$f^{3}(x)=\frac{2'(2-x)^{3}-2((2-x)^{3})'}{(2-x)^{6}}=\frac{6}{(2-x)^{4}}=\frac{3!}{(2-x)^{3+1}}$$
$$f^{4}(x)=\frac{24}{(2-x)^{5}}=\frac{4!}{(2-x)^{4+1}}$$
Following the pattern it follows:
$$f^{n}(x)=\frac{n!}{(2-x)^{n+1}}$$
