Limits Continuity And Differentiability Question 57
Question: What is the derivative of $ {{\tan }^{-1}}( \frac{\sqrt{1+x^{2}}-1}{x} ) $ with respect to $ {{\tan }^{-1}}x $ ?
Options:
A) 0
B) $ \frac{1}{2} $
C) 1
D) x
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ y={{\tan }^{-1}}[ \frac{\sqrt{1+x^{2}}-1}{x} ] $ and $ u={{\tan }^{-1}}x $ Put $ x=\tan \theta \Rightarrow \theta ={{\tan }^{-1}}x $
Then, $ y={{\tan }^{-1}}[ \frac{\sqrt{1+{{\tan }^{2}}\theta }-1}{\tan \theta } ] $
$ ={{\tan }^{-1}}[ \frac{\sqrt{{{\sec }^{2}}\theta }-1}{\tan \theta } ] $
$ ={{\tan }^{-1}}[ \frac{\sec \theta -1}{\tan \theta } ]={{\tan }^{-1}}[ \frac{\frac{1}{\cos \theta }-1}{\frac{\sin \theta }{\cos \theta }} ] $
$ ={{\tan }^{-1}}[ \frac{1-\cos \theta }{\sin \theta } ]={{\tan }^{-1}}[ \frac{2{{\sin }^{2}}\frac{\theta }{2}}{2\sin \frac{\theta }{2},\cos \frac{\theta }{2}} ] $
$ ( \begin{aligned} & \because 1-\cos \theta =2{{\sin }^{2}}\frac{\theta }{2}and \\ & \sin x=2\sin \frac{x}{2}\cdot \cos \frac{x}{2} \\ \end{aligned} ) $
$ ={{\tan }^{-1}}[ \tan \frac{\theta }{2} ] $
$ \Rightarrow y=\frac{\theta }{2}\Rightarrow y=\frac{{{\tan }^{-1}}x}{2}[\because \theta ={{\tan }^{-1}}x] $
$ \Rightarrow y=\frac{u}{2};\frac{dy}{du}=\frac{1}{2} $
BETA
