Q. Let $f(x) = e^{x},g(x) = \sin^{- 1}{x\ and\ h(x) = f\left\lbrack g(x) \right\rbrack,\ then\ \frac{h^{'}(x)}{h(x)}}$ is equal to
A) $e^{\sin^{- 1}x}$
B) $\frac{1}{\sqrt{1 - x^{2}}}$
C) $\sin^{- 1}x$
D) $\frac{1}{(1 - x^{2})}$
Explanation
(b)
$f(x) = e^{x}and\ g(x) = \sin^{- 1}x$
and $h(x) = f\lbrack g(x)\rbrack$
$\Rightarrow h(x) = f(\sin^{- 1}{x) = e^{\sin^{- 1}x}}$
$\therefore h(x) = e^{\sin^{- 1}x}$
$\Rightarrow h^{'}(x) = e^{\sin^{- 1}x}\frac{1}{\sqrt{1 - x^{2}}}$
$\Rightarrow \frac{h^{'}(x)}{h(x)} = \frac{1}{\sqrt{1 - x^{2}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.