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If $y=(\tan x)^{\sin x}$, then $\frac{d y}{d x}$ is equal to
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Verified Answer
The correct answer is:
$(\tan x)^{\sin x}\{\sec x+(\cos x)(\log (\tan x))\}$
We have,
$y=(\tan x)^{\sin x} \Rightarrow y=e^{\sin x \log \tan x}$
$\frac{d y}{d x}=e^{\sin x \log \tan x}\left(\frac{\sin x}{\tan x} \sec ^2 x+\cos x \log \tan x\right)$
$\Rightarrow \frac{d y}{d x}=(\tan x)^{\sin x}(\sec x+\cos x \log \tan x)$
$y=(\tan x)^{\sin x} \Rightarrow y=e^{\sin x \log \tan x}$
$\frac{d y}{d x}=e^{\sin x \log \tan x}\left(\frac{\sin x}{\tan x} \sec ^2 x+\cos x \log \tan x\right)$
$\Rightarrow \frac{d y}{d x}=(\tan x)^{\sin x}(\sec x+\cos x \log \tan x)$
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