A) \[{{e}^{{{\sin }^{-1}}x}}\]
B) \[1/\sqrt{1-{{x}^{2}}}\]
C) \[{{\sin }^{-1}}x\]
D) \[1/\,(1-{{x}^{2}})\]
Correct Answer: B
Solution :
\[f(x)={{e}^{x}}\] and \[g(x)={{\sin }^{-1}}x\] and \[h(x)=f(g(x))\] Þ \[h(x)\]=\[f({{\sin }^{-1}}x)={{e}^{{{\sin }^{-1}}x}}\] \ \[h\,(x)={{e}^{{{\sin }^{-1}}}}x\] Þ \[{h}'(x)={{e}^{{{\sin }^{-1}}x}}\,.\,\frac{1}{\sqrt{1-{{x}^{2}}}}\] Þ \[\frac{{h}'(x)}{h(x)}=\frac{1}{\sqrt{1-{{x}^{2}}}}\].
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