Let $f\left( x \right) = {e^x},{\rm{g}}\left( x \right) = {\sin ^{ - 1}}x\,$ and $h\left( x \right) = f\left[ {{\rm{g}}\left( x \right)} \right],$ then $\frac{{h'\left( x \right)}}{{h\left( x \right)}}$ is equal to

$\frac{1}{{\left( {1 - {x^2}} \right)}}$

JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

Let

f\left( x \right) = {e^x},\mathrm{g}\left( x \right) = {\sin ^{ - 1}}x\,

and

h\left( x \right) = f\left[ {\mathrm{g}\left( x \right)} \right],

then

\frac{{h'\left( x \right)}}{{h\left( x \right)}}

is equal to

View answer and explanation

Correct option: B

\frac{1}{{\sqrt {1 - {x^2}} }}

Option B (

\frac{1}{{\sqrt {1 - {x^2}} }}

) is correct because after finding h(x) and differentiating, the expression

\frac{{h'(x)}}{{h(x)}}

simplifies to

\frac{1}{{\sqrt {1 - {x^2}} }}