Differentiation Question 289
Question: If $ y={{\sin }^{-1}}(x\sqrt{1-x}+\sqrt{x}\sqrt{1-x^{2})}, $ then $ \frac{dy}{dx}= $
[Roorkee 1981; MP PET 2004]
Options:
A) $ \frac{-2x}{\sqrt{1-x^{2}}}+\frac{1}{2\sqrt{x-x^{2}}} $
B) $ \frac{-1}{\sqrt{1-x^{2}}}-\frac{1}{2\sqrt{x-x^{2}}} $
C) $ \frac{1}{\sqrt{1-x^{2}}}+\frac{1}{2\sqrt{x-x^{2}}} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Putting $ x=\sin A $ and $ \sqrt{x}=\sin B $
$ y={{\sin }^{-1}}(\sin A\sqrt{1-{{\sin }^{2}}B}+\sin B\sqrt{1-{{\sin }^{2}}A}) $
$ ={{\sin }^{-1}}[\sin (A+B)]=A+B={{\sin }^{-1}}x+{{\sin }^{-1}}\sqrt{x} $
Therefore $ \frac{dy}{dx}=\frac{1}{\sqrt{1-x^{2}}}+\frac{1}{2\sqrt{x-x^{2}}} $ .
BETA
