Let f(x)=ex, g(x)=sin−1x and h(x)=f(g(x)), then h'(x)h(x)=

# Let $f\left(x\right)={e}^{x},$ $g\left(x\right)={\mathrm{sin}}^{-1}x$ and $h\left(x\right)=f\left(g\left(x\right)\right),$ then $\frac{h\text{'}\left(x\right)}{h\left(x\right)}=$

1. A

${e}^{{\mathrm{sin}}^{-1}x}$

2. B

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

3. C

${\mathrm{sin}}^{-1}x$

4. D

$\frac{1}{\left(1-{x}^{2}\right)}$

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### Solution:

$⇒h\left(x\right)=f\left({\mathrm{sin}}^{-1}x\right)={e}^{{\mathrm{sin}}^{-1}x}$

$\therefore h\left(x\right)={e}^{{\mathrm{sin}}^{-1}}x⇒{h}^{\mathrm{\prime }}\left(x\right)={e}^{{\mathrm{sin}}^{-1}x}\cdot \frac{1}{\sqrt{1-{x}^{2}}}⇒\frac{{h}^{\mathrm{\prime }}\left(x\right)}{h\left(x\right)}=\frac{1}{\sqrt{1-{x}^{2}}}\text{.}$

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