Question
Differentiate the following function with respect to x:$(\log\text{x})^{x} + \text{x}^{\log\text{x}}.$

Answer

Let $y =(\log x)^x + x ^{logx}$​​​​​​​
$\Rightarrow$ y =u+v where $u=(\log x)^x , v = x^{\log x}​​​​​​​$​​​​​​​
$\Rightarrow\frac{\text{dy}}{\text{dx}} = \frac{\text{du}}{\text{dx}} + \frac{\text{dv}}{\text{dx}}$ - - - - (i)
Now $u=(\log x)^x​​​​​​​$​​​​​​​
Taking logarithm of both sides, we get
logu= x. log(log x)
Differentiating both sides w.r.t.x, we get
$\frac{1}{\text{u}}.\frac{\text{du}}{\text{dx}} =\text{x}.\frac{1}{\log\text{x}}.\frac{1}{\text{x}} + \log(\log\text{x} )\Rightarrow\frac{\text{du}}{\text{dx}} = \text{u}\left\{\frac{1}{\log\text{x}} + \log(\log\text{x})\right\}$
$\Rightarrow\frac{\text{du}}{\text{dx}} = (\log\text{x})^{x}\left\{\frac{1}{\log\text{x}} + \log(\log\text{x})\right\}$ - - - - (ii)
Again $v = x^{\log x}​​​​​​​$​​​​​​​
Taking logarithm of both sides , we get
log $v = \log x^{\log x}​​​​​​​$​​​​​​​
$\Rightarrow\log\text{v} = \log\text{x}.\log\text{x}\Rightarrow\log v = ( \log\text{x})^{2}$
Differentiating both sides w.r.t.x, we get
$\frac{1}{v}\frac{\text{dv}}{\text{dx}} = 2 \log\text{x}.\frac{1}{\text{x}}$
$\Rightarrow\frac{\text{dv}}{\text{dx}} = 2 \text{x}^{log\text{ x}}.\frac{\log\text{x}}{\text{x}}$ - - - - - (iii)
Putting $\frac{\text{du}}{\text{dx}}\text{ and } \frac{\text{dv}}{\text{dx}}$ from (ii) and (iii) in (i) we get
$\frac{\text{dy}}{\text{dx}} = ( \log\text{x})^{x}\left\{\frac{1}{\log\text{x}} + \log(\log\text{x})\right\} + 2 \frac{\log\text{x.x}^{\log\text{x}}}{\text{x}}.$

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