Editor
Editor
 Sep 18, 2022
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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: 
$(n-1)(n)(ax+b)^{n-2}$
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NETNUST Entry TestMathematicsDifferentiation