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

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

E

Editor

Sep 18, 2022

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

Difficulty: Easy

A:

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

B:

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

C:

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

D:

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

ID: 63273db8ca83163ac3a10173