Differentiation Question 264
Question: If $ y={{\tan }^{-1}}( \frac{\sqrt{a}-\sqrt{x}}{1+\sqrt{ax}} ) $ , then $ \frac{dy}{dx}= $
[AI CBSE 1988]
Options:
A) $ \frac{1}{2(1+x)\sqrt{x}} $
B) $ \frac{1}{(1+x)\sqrt{x}} $
C) $ -\frac{1}{2(1+x)\sqrt{x}} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ y={{\tan }^{-1}}\sqrt{a}-{{\tan }^{-1}}\sqrt{x} $
Differentiating w.r.t. x, we get, $ \frac{dy}{dx}=-\frac{1}{(1+x)}.\frac{1}{2\sqrt{x}} $ .
BETA
