Differentiation Question 342
Question: If $ y=\sin (2{{\sin }^{-1}}x), $ then $ \frac{dy}{dx}= $
[AI CBSE 1983]
Options:
A) $ \frac{2-4x^{2}}{\sqrt{1-x^{2}}} $
B) $ \frac{2+4x^{2}}{\sqrt{1-x^{2}}} $
C) $ \frac{2-4x^{2}}{\sqrt{1+x^{2}}} $
D) $ \frac{2+4x^{2}}{\sqrt{1+x^{2}}} $
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ x=\sin \theta \Rightarrow 2{{\sin }^{-1}}x=2\theta $
Therefore $ y=\sin 2\theta $
Therefore $ \frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{2\cos 2\theta }{\cos \theta } $
$ =\frac{2(1-2{{\sin }^{2}}\theta )}{\sqrt{1-{{\sin }^{2}}\theta }}=\frac{2-4x^{2}}{\sqrt{1-x^{2}}} $ .
BETA
