A) 0
B) \[\frac{1}{2}\]
C) ? 1
D) \[-\frac{1}{4}\]
Correct Answer: D
Solution :
\[y={{\tan }^{-1}}\left( \frac{\sqrt{x}-x}{1+{{x}^{3/2}}} \right)={{\tan }^{-1}}(\sqrt{x})-{{\tan }^{-1}}(x)\] Þ\[{y}'=\frac{1}{1+x}.\frac{1}{2\sqrt{x}}-\frac{1}{1+{{x}^{2}}}\Rightarrow {y}'(1)=\frac{1}{2}.\frac{1}{2}-\frac{1}{2}=\frac{-1}{4}\].
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