Differentiation Question 29
Question: $ \frac{d}{dx}[ {{\sin }^{2}}{{\cot }^{-1}}{ \sqrt{\frac{1-x}{1+x}} } ] $ equals
[MP PET 2002]
Options:
A) $ -1 $
B) $ \frac{1}{2} $
C) $ -\frac{1}{2} $
D) 1
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ y={{\sin }^{2}}{{\cot }^{-1}}{ \sqrt{\frac{1-x}{1+x}} } $
Put $ x=\cos \theta \Rightarrow \theta ={{\cos }^{-1}}x $
Therefore $ y={{\sin }^{2}}{{\cot }^{-1}}{ \sqrt{\frac{1-\cos \theta }{1+\cos \theta }} }={{\sin }^{2}}{{\cot }^{-1}}( \tan \frac{\theta }{2} ) $
Therefore $ y={{\sin }^{2}}( \frac{\pi }{2}-\frac{\theta }{2} ) $
Therefore $ \frac{dy}{d\theta }=2\sin ( \frac{\pi }{2}-\frac{\theta }{2} ).\cos ( \frac{\pi }{2}-\frac{\theta }{2} )( -\frac{1}{2} ) $
Therefore $ \frac{dy}{d\theta }=-\frac{\sin (\pi -\theta )}{2}=-\frac{\sin \theta }{2}=\frac{-1}{2}\sqrt{1-x^{2}} $
Therefore $ \frac{dy}{dx}=\frac{dy}{d\theta }.\frac{d\theta }{dx}=\frac{-1}{2}\sqrt{1-x^{2}}.\frac{d}{dx}({{\cos }^{-1}}x)=\frac{1}{2} $ .
BETA
