Question

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

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

Answer 1

Answer: (B)

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