Differentiation Question 35
Question: If $ y={{\sin }^{-1}}\sqrt{1-x^{2}} $ , then $ dy/dx= $
[AISSE 1987]
Options:
A) $ \frac{1}{\sqrt{1-x^{2}}} $
B) $ \frac{1}{\sqrt{1+x^{2}}} $
C) $ -\frac{1}{\sqrt{1-x^{2}}} $
D) $ -\frac{1}{\sqrt{x^{2}-1}} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ y={{\sin }^{-1}}(\sqrt{1-x^{2}}) $
Let $ \sqrt{1-x^{2}}=\sin \theta \Rightarrow 1-x^{2}={{\sin }^{2}}\theta $
Therefore $ x^{2}=1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta $
$ \therefore x=\cos \theta $ or $ \theta ={{\cos }^{-1}}x $
$ \Rightarrow y={{\cos }^{-1}}x $
Differentiating w.r.t. x of y, we get $ \frac{dy}{dx}=-\frac{1}{\sqrt{1-x^{2}}} $ .
BETA
