Question

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

• A. $e^{sin^{-1}\,x}$
• B. $\frac{1}{\sqrt{1-x^{2}}}$
• C. $sin^{-1}\,x$
• D. $\frac{1}{\left(1-x^{2}\right)}$

Answer 1

Answer: (B)

$f(x) = e^x$ and $g(x)$
$ = sin^{-1}x$ and $h(x) =f[g(x)]$
$\Rightarrow h\left(x\right)=f \left(sin^{-1}\,x\right)=e^{sin^{-1}\,x}$
$\therefore h'\left(x\right)=e^{sin^{-1}\,x}\left(\frac{1}{\sqrt{1-x^{2}}}\right)$
$=h\left(x\right)\cdot\frac{1}{\sqrt{1-x^{2}}}$
$\Rightarrow \frac{h'\left(x\right)}{h\left(x\right)}=\frac{1}{\sqrt{1-x^{2}}}$