Differentiation Question 401
Question: If $ y={{\tan }^{-1}}\sqrt{\frac{a-x}{a+x}} $ , then $ \frac{dy}{dx}= $
Options:
A) $ {{\cos }^{-1}}\frac{x}{a} $
B) $ -{{\cos }^{-1}}\frac{x}{a} $
C) $ \frac{1}{2}{{\cos }^{-1}}\frac{x}{a} $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
Putting $ x=a\cos \theta \Rightarrow \theta ={{\cos }^{-1}}\frac{x}{a} $
$ y={{\tan }^{-1}}\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}=\frac{\theta }{2}=\frac{1}{2}{{\cos }^{-1}}\frac{x}{a} $
Therefore $ \frac{dy}{dx}=-\frac{1}{2}\frac{1}{\sqrt{1-\frac{x^{2}}{a^{2}}}}.\frac{1}{a}=-\frac{1}{2\sqrt{a^{2}-x^{2}}} $ .
BETA
