Differentiation Question 163
Question: The differential coefficient of $ {{\tan }^{-1}}\sqrt{x} $ with respect to $ \sqrt{x} $ is
[MP PET 1987]
Options:
A) $ \frac{1}{\sqrt{1+x}} $
B) $ \frac{1}{2x\sqrt{1+x}} $
C) $ \frac{1}{2\sqrt{x(1+x)}} $
D) $ \frac{1}{1+x} $
Show Answer
Answer:
Correct Answer: D
Solution:
Let $ y_1={{\tan }^{-1}}\sqrt{x} $ and $ y_2=\sqrt{x} $
Differentiating w.r.t. x of $ y_1 $ and $ y_2 $ , we get $ \frac{dy_1}{dx}=\frac{1}{(1+x)}.\frac{1}{2\sqrt{x}} $ and $ \frac{dy_2}{dx}=\frac{1}{2\sqrt{x}} $
Hence $ \frac{dy_1}{dy_2}=\frac{1}{1+x} $ .
BETA
