Question

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

• A. $e^{{\sin }^{-1}x}$
• B. $\frac{1}{\sqrt{1-x^2}}$
• C. ${\sin }^{-1}x$
• D. $\frac{1}{1-x^2}$

Answer 1

The correct option is B

$\frac{1}{\sqrt{1-x^2}}$

Explanation for the correct option.

Step 1. Find the values of $h\left(x\right)$and $h\text{'}\left(x\right)$.

For the functions $f\left(x\right)=e^x,g\left(x\right)={\sin }^{-1}x$, the function $h\left(x\right)=f\left(g\left(x\right)\right)$ is defined as:

$\begin{array}{rcl}h\left(x\right)& =& f\left(g\left(x\right)\right)\\ & =& f\left({\sin }^{-1}x\right)\\ & =& e^{{\sin }^{-1}x}\end{array}$

Now, differentiate the function $h\left(x\right)=e^{{\sin }^{-1}x}$.

$\begin{array}{rcl}h\text{'}\left(x\right)& =& \frac{d}{dx}\left(e^{{\sin }^{-1}x}\right)\\ & =& e^{{\sin }^{-1}x}\frac{1}{\sqrt{1-x^2}}\\ & =& \frac{e^{{\sin }^{-1}x}}{\sqrt{1-x^2}}\end{array}$

Step 2. Find the value of $\frac{h\text{'}\left(x\right)}{h\left(x\right)}$.

The value of $\frac{h\text{'}\left(x\right)}{h\left(x\right)}$ is given as:

$\begin{array}{rcl}\frac{h\text{'}\left(x\right)}{h\left(x\right)}& =& \frac{{\displaystyle \frac{e^{{\sin }^{-1}x}}{\sqrt{1-x^2}}}}{e^{{\sin }^{-1}x}}\\ & =& \frac{1}{\sqrt{1-x^2}}\end{array}$

Hence, the correct option is B.