Limits Continuity And Differentiability Question 96
Question: The derivative of $ {{\sin }^{-1}}( \frac{2x}{1+x^{2}} ) $ with respect to $ {{\cos }^{-1}}[ \frac{1-x^{2}}{1+x^{2}} ] $ is equal to:
Options:
A) 1
B) -1
C) 2
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ s={{\sin }^{-1}}( \frac{2x}{1+x^{2}} ) $ and t
$ ={{\cos }^{-1}}( \frac{1-x^{2}}{1+x^{2}} ) $
We have to find out $ \frac{ds}{dt}; $ putting $ x=\tan \theta , $ we get
$ s={{\sin }^{-1}}[ \frac{2\tan \theta }{1+{{\tan }^{2}}\theta } ]={{\sin }^{-1}}(\sin 2\theta )=2\theta $
$ =2{{\tan }^{-1}}x $
$ \therefore \frac{ds}{dx}=\frac{2}{1+x^{2}} $
and $ t={{\cos }^{-1}}( \frac{1-x^{2}}{1+x^{2}} )={{\cos }^{-1}}( \frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } ) $
$ ={{\cos }^{-1}}(cos2\theta )=2\theta =2ta{n^{-1}}x $
$ \therefore \frac{dt}{dx}=\frac{2}{1+x^{2}} $
$ \therefore \frac{ds}{dt}=\frac{ds/dx}{dt/dx}=\frac{2}{1+x^{2}}\times \frac{1+x^{2}}{2}=1 $
BETA
