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

1 Medium

1 of 1

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 it is the result of differentiating the composite function h(x) and simplifying the expression.