Differentiation Question 102
Question: The differential coefficient of $ {{\tan }^{-1}}\frac{2x}{1-x^{2}} $ w.r.t. $ {{\sin }^{-1}}\frac{2x}{1+x^{2}} $ is
[Roorkee 1966; BIT Ranchi 1996; Karnataka CET 1994; MP PET 1999; UPSEAT 1999, 2001]
Options:
A) 1
B) - 1
C) 0
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ y_1={{\tan }^{-1}}\frac{2x}{1-x^{2}} $ and $ y_2={{\sin }^{-1}}\frac{2x}{1+x^{2}} $
Differentiating w.r.t. x of $ y_1 $ and $ y_2 $ , we get $ \frac{dy_1}{dx}=\frac{d}{dx}[ {{\tan }^{-1}}\frac{2x}{1-x^{2}} ] $
Putting $ x=\tan \theta $ ,
$ \therefore y_1={{\tan }^{-1}}\tan 2\theta =2\theta =2{{\tan }^{-1}}x $
and $ y_2={{\sin }^{-1}}\sin 2\theta =2{{\tan }^{-1}}x $
Again $ \frac{dy_1}{dx}=\frac{d}{dx}[2{{\tan }^{-1}}x]=\frac{2}{1+x^{2}} $
…..(i) and $ \frac{dy_2}{dx}=\frac{d}{dx}[2{{\tan }^{-1}}x]=\frac{2}{1+x^{2}} $
…..(ii) Hence $ \frac{dy_1}{dy_2}=1 $ .
BETA
