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
$\log u= x. \log(\log x)$
Differentiating both sides $\text{w.r.t.x},$ 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 $\text{w.r.t.x},$ 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 $\frac{\text{du}}{\text{dx}}$ 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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