A) \[\frac{1}{\sqrt{1+x}}\]
B) \[\frac{1}{2x\sqrt{1+x}}\]
C) \[\frac{1}{2\sqrt{x(1+x)}}\]
D) \[\frac{1}{1+x}\]
Correct Answer: D
Solution :
Let \[{{y}_{1}}={{\tan }^{-1}}\sqrt{x}\]and \[{{y}_{2}}=\sqrt{x}\] Differentiating w.r.t. x of\[{{y}_{1}}\]and \[{{y}_{2}}\], we get \[\frac{d{{y}_{1}}}{dx}=\frac{1}{(1+x)}.\frac{1}{2\sqrt{x}}\]and \[\frac{d{{y}_{2}}}{dx}=\frac{1}{2\sqrt{x}}\] Hence \[\frac{d{{y}_{1}}}{d{{y}_{2}}}=\frac{1}{1+x}\].
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