A) \[\frac{1}{2}\]
B) \[\frac{2}{3}\]
C) \[1\]
D) \[2\]
Correct Answer: D
Solution :
\[f\left( x \right)={{\tan }^{-1}}\left( \frac{\sin x-\cos x}{\sin x+\cos x} \right)\] \[={{\tan }^{-1}}\left( \frac{\tan x-1}{\tan x+1} \right)={{\tan }^{-1}}\left( \tan \left( x-\frac{\pi }{4} \right) \right)\] \[\because \]\[=x-\frac{\pi }{4}\in \left( -\frac{\pi }{4},\frac{\pi }{4} \right)\] \[\therefore \]\[=f\left( x \right)=x-\frac{\pi }{4}\] \[\Rightarrow \]its derivative w.r.t. \[\frac{x}{2}\] is \[\frac{1}{1/2}=2\]
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