Differentiation Question 347
Question: If $ y={{\tan }^{-1}}\frac{x}{1+\sqrt{1-x^{2}}}+\sin { 2{{\tan }^{-1}}\sqrt{( \frac{1-x}{1+x} )} } $ ,then $ \frac{dy}{dx} $ =
Options:
A) $ \frac{x}{\sqrt{1-x^{2}}} $
B) $ \frac{1-2x}{\sqrt{1-x^{2}}} $
C) $ \frac{1-2x}{2\sqrt{1-x^{2}}} $
D) $ \frac{1}{1+x^{2}} $
Show Answer
Answer:
Correct Answer: C
Solution:
Put $ x=\cos \theta $
$ \therefore y={{\tan }^{-1}}\frac{\cos \theta }{1+\sin \theta }+\sin [ 2{{\tan }^{-1}}\sqrt{( \frac{1-\cos \theta }{1+\cos \theta } )} ] $
$ ={{\tan }^{-1}}\frac{\sin \varphi }{1+\cos \varphi }+\sin [ 2{{\tan }^{-1}}\tan ( \frac{\theta }{2} ) ] $ , {where $ \varphi =90{}^\circ -\theta $ } $ ={{\tan }^{-1}}\tan ( \frac{\varphi }{2} )+\sin ( 2.\frac{\theta }{2} )=( \frac{\varphi }{2} )+\sin \theta $
$ =\frac{\pi }{4}-\frac{\theta }{2}+\sqrt{1-{{\cos }^{2}}\theta }=\frac{\pi }{4}-\frac{1}{2}{{\cos }^{-1}}x+\sqrt{1-x^{2}} $
$ \therefore \frac{dy}{dx}=\frac{1}{2}\frac{1}{\sqrt{1-x^{2}}}+\frac{1}{2\sqrt{1-x^{2}}}(-2x)=\frac{1-2x}{2\sqrt{1-x^{2}}} $
BETA
