Differentiation Question 381
Question: If $ y={{\sin }^{-1}}\sqrt{(1-x)}+{{\cos }^{-1}}\sqrt{x} $ , then $ \frac{dy}{dx}= $
Options:
A) $ \frac{1}{\sqrt{x(1-x)}} $
B) $ \frac{-1}{\sqrt{x(1-x)}} $
C) $ \frac{1}{\sqrt{x(1+x)}} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ {{\sin }^{-1}}\sqrt{1-x}={{\sin }^{-1}}\sqrt{1-{{(\sqrt{x})}^{2}}}={{\cos }^{-1}}\sqrt{x} $
$ \therefore y=2{{\cos }^{-1}}\sqrt{x} $ or $ \frac{dy}{dx}=2.\frac{-1}{\sqrt{1-x}}.\frac{1}{2\sqrt{x}} $ etc. Aliter: $ y={{\sin }^{-1}}\sqrt{1-x}+{{\cos }^{-1}}\sqrt{x} $
Therefore $ \frac{dy}{dx}=\frac{1}{\sqrt{1-1+x}}.\frac{-1}{2\sqrt{1-x}}-\frac{1}{\sqrt{1-x}}.\frac{1}{2\sqrt{x}} $
$ =\frac{-2}{2\sqrt{x}\sqrt{1-x}}=\frac{-1}{\sqrt{x}\sqrt{1-x}} $ .
BETA
