Differentiation Question 112
Question: $ \frac{d}{dx}{ {{\cos }^{-1}}( \frac{1-x^{2}}{1+x^{2}} ) }= $
[AISSE 1984]
Options:
A) $ \frac{1}{1+x^{2}} $
B) $ -\frac{1}{1+x^{2}} $
C) $ -\frac{2}{1+x^{2}} $
D) $ \frac{2}{1+x^{2}} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \frac{d}{dx}{ {{\cos }^{-1}}( \frac{1-x^{2}}{1+x^{2}} ) } $
Let $ \frac{1-x^{2}}{1+x^{2}}=\cos \theta $
Therefore $ 1-x^{2}=(1+x^{2})\cos \theta $
Therefore $ -x^{2}(1+\cos \theta )=\cos \theta -1 $
Therefore $ x^{2}=\frac{1-\cos \theta }{1+\cos \theta }=\frac{2{{\sin }^{2}}\frac{\theta }{2}}{2{{\cos }^{2}}\frac{\theta }{2}}={{\tan }^{2}}\frac{\theta }{2} $
or $ x=\tan \frac{\theta }{2} $ or $ \theta =2{{\tan }^{-1}}x $
So, $ \frac{d}{dx}[\theta ]=\frac{d}{dx}[2{{\tan }^{-1}}x]=\frac{2}{1+x^{2}} $ .
BETA
