A) \[{{(\sin x)}^{\tan x}}[1+{{\sec }^{2}}x\log \sin x]\]
B) \[{{(\sin x)}^{\tan x}}[1-{{\sec }^{2}}x\log \sin x]\]
C) \[{{(\tan x)}^{\sin x}}[\log {{\sec }^{2}}x\log \sin x]\]
D) None of the above
Correct Answer: A
Solution :
\[\frac{d}{dx}{{(\sin x)}^{\tan x}}\] \[={{(\sin x)}^{\tan x}}\left[ \tan x.\frac{\cos x}{\sin x}+{{\sec }^{2}}x\log (\sin x) \right]\] \[\left[ \because \frac{d}{dx}[f{{(x)}^{g(x)}} \right.\] \[\left. f{{(x)}^{g(x)}}\left( g(x)\frac{f'(x)}{f(x)}+g'(x)\log f(x) \right) \right]\] \[={{(\sin x)}^{\tan x}}\left[ \frac{\tan x}{\tan x}+{{\sec }^{2}}x.\log (\sin x) \right]\] \[={{(\sin x)}^{\tan x}}[1+{{\sec }^{2}}x.\log (\sin x)]\]
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