Differentiation Question 11
Question: If $ y={{(\sin x)}^{\tan x}} $ , then $ \frac{dy}{dx} $ is equal to
[IIT 1994; RPET 1996]
Options:
A) $ {{(\sin x)}^{\tan x}}.(1+{{\sec }^{2}}x.\log \sin x) $
B) $ \tan x.{{(\sin x)}^{\tan x-1}}.\cos x $
C) $ {{(\sin x)}^{\tan x}}.{{\sec }^{2}}x.\log \sin x $
D) $ \tan x.{{(\sin x)}^{\tan x-1}} $
Show Answer
Answer:
Correct Answer: A
Solution:
Given $ y={{(\sin x)}^{\tan x}} $ ; $ \log y=\tan x.\log \sin x $
Differentiate with respect to x, $ \frac{1}{y}.\frac{dy}{dx}=\tan x.\cot x+\log \sin x.{{\sec }^{2}}x $
$ \frac{dy}{dx}={{(\sin x)}^{\tan x}}[1+\log \sin x.{{\sec }^{2}}x] $ .
BETA
