Differentiation Question 262
The derivative of $ {{\sin }^{2}}x $ with respect to $ x $ is
[DCE 2002]
Options:
A) $ {{\tan }^{2}}x $
B) $ \tan x $
C) $ -\tan x $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
Let $ y={{\sin }^{2}}x $ and $ z={{\cos }^{2}}x $
Therefore $ \frac{dy}{dx}=\sin 2x $ and $ \frac{dz}{dx}=-\sin 2x $ , \ $ \frac{dy}{dz}=\frac{dy}{dx}\cdot\frac{dx}{dz}=\frac{\sin 2x}{-\sin 2x}=-1 $ .
BETA
