If y=e^ x +( x)^ x then find d y d x

Question

If $y=e^{\tan x}+(\log x)^{\tan x}$ then find $\frac{d y}{d x}$

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Answer

Let $y = e^{\tan x}+ (\log x)^{\tan x}$
Put $u=e^{ tanx}$ and $v=(logx)^{tanx}$
$y=u+v$
$\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}......(1)$
$u=e^{\tan x}$
Taking logarithm on both sides, we get
$\log u = \tan x.\log e = \tan x$
Differentiating $w. r. t. x,$ we get
$\frac{1}{u} \frac{d u}{d x}=\sec ^2 x$
$\therefore \frac{d u}{d x}=u \cdot \sec ^2 x$
$\therefore \frac{d u}{d x}=e^{\tan x} \cdot \sec ^2 x......(2)$
$v=(\log x)^{\tan x}$
Taking logarithm on both sides, we get
$\log v = \tan x.\log (\log x)$
Differentiating $w.r.t. x,$ we get
$\frac{1}{v} \frac{d v}{d x}=\tan x \frac{.1}{\log x} \frac{1}{x}+\log (\log x) \sec ^2 x$
$\frac{d v}{d x}=v\left[\frac{\tan x}{x \log x}+\log (\log x) \sec ^2 x\right]$
$=(\log x)^{\tan x}\left[\frac{\tan x}{x \log x}+\log (\log x) \sec ^2 x\right]......(3)$
From $(1), (2)$ and $(3),$ we get
$\frac{d y}{d x}=e^{\tan x} \cdot \sec ^2 x+(\log x)^{\tan x}\left[\frac{\tan x}{x \log x}+\log (\log x) \sec ^2 x\right]$

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