Applications Of Derivatives Question 20
Question: What is the slope of the tangent to the curve $ y={{\sin }^{-1}}({{\sin }^{2}}x)atx=0 $ -
Options:
A) 0
B) 1
C) 2
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ y={{\sin }^{-1}}({{\sin }^{2}}x) $
$ \frac{dy}{dx}=\frac{2\sin x\cos x}{\sqrt{1-{{\sin }^{4}}x}}\Rightarrow \frac{dy}{dx}=\frac{\sin 2x}{\sqrt{1-{{\sin }^{4}}x}} $
$ atx=0,\frac{dy}{dx}=0 $
BETA
