A) \[\frac{1}{2}\]
B) 1
C) 2
D) \[\frac{3}{2}\]
Correct Answer: A
Solution :
\[y={{\tan }^{-1}}\left[ \frac{x}{1+\sqrt{1-{{x}^{2}}}} \right]\] Put \[x=\sin \theta \] Þ \[y={{\tan }^{-1}}\left[ \frac{\sin \theta }{1+\cos \theta } \right]\,=\,{{\tan }^{-1}}\,\tan \frac{\theta }{2}=\frac{\theta }{2}\] Þ \[y=\frac{1}{2}{{\sin }^{-1}}x\] and let \[z={{\sin }^{-1}}x\] Hence \[\frac{dy}{dz}=\frac{\left( \frac{dy}{dx} \right)}{\left( \frac{dz}{dx} \right)}\]=\[\frac{\frac{1}{2}\frac{d}{dx}{{\sin }^{-1}}x}{\frac{d}{dx}{{\sin }^{-1}}x}\] = \[f''({{e}^{x}}).{{e}^{2x}}+f'({{e}^{x}}).{{e}^{x}}\].
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