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Editor
 Sep 18, 2022
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If $f(x)=\frac{1}{\sqrt{x^{2}-1}}$, then f'(x) equals:
Difficulty: Easy
A: 
$\frac{2}{\sqrt{x^{2}-1}}$
B: 
$-\frac{x}{\sqrt{(x^{2}-1)^{3}}}$
C: 
$\frac{x}{\sqrt{x^{2}-1}}$
D: 
$\frac{3x}{\sqrt{x^{2}-1}}$
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NETNUST Entry TestMathematicsDifferentiation