Differentiation Question 10
Question: $ y={{(\tan x)}^{{{(\tan x)}^{\tan x}}}}, $ then at $ x=\frac{\pi }{4} $ , the value of $ \frac{dy}{dx}= $
[WB JEE 1990]
Options:
A) 0
B) 1
C) 2
D) None of these
Show Answer
Answer:
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.[ \log \tan x+1+\frac{1}{\tan x\log \tan x} ] $
$ =y{{(\tan x)}^{\tan x}}\log \tan x.{{\sec }^{2}}x $
$ [ (\log \tan x+1)+\frac{1}{\tan x\log \tan x} ] $
$ =y{{(\tan x)}^{\tan x}}{{\sec }^{2}}x[ \log \tan x(\log \tan x+1)+\cot x ] $
Now at $ x=\frac{\pi }{4},\ \ y=1,\ \ \log \tan ( \frac{\pi }{4} )=\log 1=0 $
$ \therefore \frac{dy}{dx}=1.1.2[0+1]=2 $ .
BETA
