A) \[\frac{1}{2}\]
B) \[-\frac{1}{2}\]
C) 1
D) 0
Correct Answer: A
Solution :
Let \[y={{\tan }^{-1}}\sqrt{\frac{1-{{x}^{2}}}{1+{{x}^{2}}}}\]and \[z={{\cos }^{-1}}({{x}^{2}})\] Put \[{{x}^{2}}=\cos 2\theta \];\[\therefore y={{\tan }^{-1}}\sqrt{\frac{2{{\sin }^{2}}\theta }{2{{\cos }^{2}}\theta }}=\theta \]and \[z=2\theta \] \[\therefore \frac{dy}{dz}=\frac{dy/d\theta }{dz/d\theta }=\frac{1}{2}\].
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