Differentiate the function ( x x )^x + ( x x )^ 1 x w.r.t. x.

Question

Differentiate the function ${\left( {x\cos x} \right)^x} + {\left( {x\sin x} \right)^{\frac{1}{x}}}$ w.r.t. x.

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Answer

Let $y = {\left( {x\cos x} \right)^x} + {\left( {x\sin x} \right)^{\frac{1}{x}}}$
Putting $u = (x \cos x)^x$ and $v = {\left( {x\sin x} \right)^{\frac{1}{x}}}$
we have $y = u + v$
$\therefore \frac{{dy}}{{dx}} = \frac{{du}}{{dx}} + \frac{{dv}}{{dx}} ....(i)$
Now $u = (x \cos x)^x$
$\Rightarrow \log u = \log {\left( {x\cos x} \right)^x} = x\log \left( {x\cos x} \right)$
$\Rightarrow \log u = x\left( {\log x + \log \cos x} \right)$
$\Rightarrow \frac{d}{{dx}}\log u = \frac{d}{{dx}}\left\{ {x\left( {\log x + \log \cos x} \right)} \right\}$
$\Rightarrow \frac{1}{u}\frac{{du}}{{dx}} = x\left[ {\frac{1}{x} + \frac{1}{{\cos x}}.\left( { - \sin x} \right)} \right] + \left( {\log x + \log \cos x} \right).1$
$\Rightarrow \frac{1}{u}\frac{{du}}{{dx}} = \left[ {1 - x\tan x + \log \left( {x\cos x} \right)} \right]$
$\Rightarrow \frac{{du}}{{dx}} = u\left[ {1 - x\tan x + \log \left( {x\cos x} \right)} \right]$
$\Rightarrow \frac{{du}}{{dx}} = {\left( {x\cos x} \right)^x}\left[ {1 - x\tan x + \log \left( {x\cos x} \right)} \right] .....(ii)$
Again $v = {\left( {x\sin x} \right)^{\frac{1}{x}}}$
$\Rightarrow \log v = \log {\left( {x\sin x} \right)^{\frac{1}{x}}} = \frac{1}{x}\log \left( {x\sin x} \right)$
$\Rightarrow \log v = \frac{1}{x}\left( {\log x + \log \sin x} \right)$
$\Rightarrow \frac{d}{{dx}}\log v = \frac{d}{{dx}}\left\{ {\frac{1}{x}\left( {\log x + \log \sin x} \right)} \right\}$
$\Rightarrow \frac{1}{v}\frac{dv}{{dx}} = \frac{1}{x}\left[ {\frac{1}{x} + \frac{1}{{\ sinx }}.\cos x} \right] + \left( {\log x + \log \sin x} \right)\left( {\frac{{ - 1}}{{{x^2}}}} \right)$
$\Rightarrow \frac{1}{v}\frac{dv}{{dx}} = \left[ {\frac{1}{{{x^2}}} + \frac{{\cot x}}{x} - \frac{{\log \left( {x\sin x} \right)}}{{{x^2}}}} \right]$
$\Rightarrow \frac{dv}{{dx}} = v\left[ {\frac{1}{{{x^2}}} + \frac{{\cot x}}{x} - \frac{{\log \left( {x\sin x} \right)}}{{{x^2}}}} \right]$
$\Rightarrow \frac{dv}{{dx}} = {\left( {x\sin x} \right)^{\frac{1}{x}}}\left[ {\frac{1}{{{x^2}}} + \frac{{\cot x}}{x} - \frac{{\log \left( {x\sin x} \right)}}{{{x^2}}}} \right]...(iii)$
Putting the values from eq. $(ii)$ and $(iii)$ in eq. $(i)$
$\frac{d}{{dx}} = {\left( {x\cos x} \right)^{^x}} \left[ {1 - x\tan x + \log \left( {x\cos x} \right)} \right] + {\left( {x\sin x} \right)^{\frac{1}{x}}}$$\left[ {\frac{1}{{{x^2}}} + \frac{{\cot x}}{x} - \frac{{\log \left( {x\sin x} \right)}}{{{x^2}}}} \right]$