If $f(x)=\frac{1}{(ax+b)^{n}}$ then f'(x) equals:

Question

Accepted Answer

$\frac{na}{(ax+b)^{n+1}}$

Answer

$nx(ax+b)^{n+1}$

Answer

$nx^{n-1}(ax+b)$

Answer

$(n-1)(n)(ax+b)^{n-2}$