Limits Continuity And Differentiability Question 94
Question: Let $ 0<x<\pi $ and y(x) be given by $ (1+\sin x)y^{3}-(\cos x)y^{2}+2(1+\sin x) $ $ y-2\cos x=0 $ . The derivative of y with respect to $ \tan \frac{x}{2} $ at $ x=\frac{\pi }{2} $ is:
Options:
A) $ \frac{1}{2} $
B) $ -\frac{1}{2} $
C) $ 2 $
D) $ -2 $
Show Answer
Answer:
Correct Answer: B
Solution:
The given eq. can be written as $ (y^{2}+2)( y-\frac{\cos x}{1+\sin x} )=0 $
$ \Rightarrow y=\frac{\cos x}{1+\sin x}[\because y^{2}+2\ne 0] $
$ =\frac{\frac{1-{{\tan }^{2}}\frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}}}{1+\frac{2\tan \frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}}}=\frac{1-t^{2}}{1+t^{2}+2t}=\frac{(1-t)(1+t)}{{{(1+t)}^{2}}} $
$ =\frac{1-t}{1+t}=\frac{2}{1+t}-1, $ where $ t=\tan \frac{x}{2} $
$ \Rightarrow y’(t)=\frac{-2}{{{(1+t)}^{2}}}.Atx=\frac{\pi }{2},t=1 $
$ \therefore y’(1)=\frac{-2}{{{(1+1)}^{2}}}=-\frac{1}{2} $
BETA
