Q. Let $f(x) = e^{x},g(x) = \sin^{- 1}xand\ h(x) = f\lbrack g(x)\rbrack,then\frac{h^{'}(x)}{h(x)}$ 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)
Given, $f(x) = e^{x},\ \ g(x) = \sin^{- 1}x$
Since, $h(x) = f\lbrack g(x)\rbrack = e^{\sin^{- 1}x}$
Now, $h^{'}(x) = \ e^{\sin x}.\frac{1}{\sqrt{1 - x^{2}}}$
$\Rightarrow h^{'}(x) = h(x).\frac{1}{\sqrt{1 - x^{2}}}$
$\Rightarrow \frac{h^{'}(x)}{h(x)} = \frac{1}{\sqrt{1 - x^{2}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.