A) 0
B) 1
C) 2
D) None of these
Correct Answer: C
Solution :
\[\log y={{(\tan x)}^{\tan x}}\log \tan x\] .....(i) Taking \[\log \]again, we get from (i) \[\log (\log y)=\tan x\log \tan x+\log (\log \tan x)\] Differentiating w.r.t. x, \[\frac{1}{\log y}.\frac{1}{y}\frac{dy}{dx}\] \[={{\sec }^{2}}x\log \tan x+\tan x.\frac{{{\sec }^{2}}x}{\tan x}+\frac{1}{\log \tan x}.\frac{1}{\tan x}.{{\sec }^{2}}x\] \[\therefore \frac{dy}{dx}=y\log y{{\sec }^{2}}x.\left[ \log \tan x+1+\frac{1}{\tan x\log \tan x} \right]\] \[=y{{(\tan x)}^{\tan x}}\log \tan x.{{\sec }^{2}}x\] \[\left[ (\log \tan x+1)+\frac{1}{\tan x\log \tan x} \right]\] \[=y{{(\tan x)}^{\tan x}}{{\sec }^{2}}x\left[ \log \tan x(\log \tan x+1)+\cot x \right]\] Now at \[x=\frac{\pi }{4},\ \ y=1,\ \ \log \tan \left( \frac{\pi }{4} \right)=\log 1=0\] \[\therefore \frac{dy}{dx}=1.1.2[0+1]=2\].
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