${(x\cos x)^x} + {(x\sin x)^{1/x}}$

Let $y = {(x\cos x)^x} + {(x\sin x)^{1/x}} = u + v,$

where$u = {(x\cos x)^x},v = {(x\sin x)^{1/x}}$

$\Rightarrow$ $\cfrac{{dy}}{{dx}} = \cfrac{{du}}{{dx}} + \cfrac{{dv}}{{dx}}$

…(i)
Now, $u = {(x\cos x)^x}.$

By taking log on both sides , we get log u $=$ x log (x cos x) ….(ii)

Differentiating (ii) on both sides w.r.t. x, we get

$\cfrac{1}{u}\cfrac{{du}}{{dx}} = x\cfrac{d}{{dx}}\log (x\cos x) + \log (x\cos x)$

$= x\left[ {\cfrac{1}{{x\cos x}}\cfrac{d}{{dx}}(x\cos x)} \right] + \log (x\cos x)$

$= x\left[ {\cfrac{1}{{x\cos x}}\left( {x\cfrac{d}{{dx}}\cos x + \cos x} \right)} \right] + \log (x\cos x)$

$= x\left[ {\cfrac{1}{{x\cos x}}\{ x( - \sin x) + \cos x\} } \right] + \log (x\cos x)$

$= \sec x(\cos x - \sin x) + \log (x\cos x)$
$= 1 - x\tan x + \log (x\cos x)$

therefore, $\cfrac{{du}}{{dx}} = {(x\cos x)^x}[1 - x\tan x + \log (x\cos x)]$ ….(iii)

Now, $v = {(x\sin x)^{1/x}}$

By taking log on both sides , we get
$\log v = \cfrac{1}{x}(x\sin x)$

…(iv)
Differentiating (iv) w.r.t.x, we get

$\cfrac{1}{v}\cfrac{{dv}}{{dx}} = \cfrac{1}{x}\cfrac{d}{{dx}}\log (x\sin x) + \log (x\sin x)\cfrac{d}{{dx}}\left( {\cfrac{1}{x}} \right)$

$= \cfrac{1}{x} \cdot \cfrac{1}{{x\sin x}}\cfrac{d}{{dx}}(x\sin x) + \log (x\sin x) \cdot \left( {\cfrac{{ - 1}}{{{x^2}}}} \right)$

$= \cfrac{1}{{{x^2}\sin x}}\left[ {x\cfrac{d}{{dx}}(\sin x) + \sin x} \right] - \cfrac{{\log (x\sin x)}}{{{x^2}}}$ =

$= \cfrac{{\cot x}}{x} + \cfrac{1}{{{x^2}}} - \cfrac{{\log (x\sin x)}}{{{x^2}}} = \cfrac{{x\cot x + 1 - \log (x\sin x)}}{{{x^2}}}$

therefore, $\cfrac{{dv}}{{dx}} = {(x\sin x)^{1/x}}\left[ {\cfrac{{x\cot x + 1 - \log (x\sin x)}}{{{x^2}}}} \right]$ …(v)

Substituting the values of (iii) and (v) in (i), we get
$\cfrac{{dy}}{{dx}} = {(x\cos x)^x}[1 - x\tan x + \log (x\cos x)] + {(x\sin x)^{1/x}}\left[ {\cfrac{{x\cot x + 1 - \log (x\sin x)}}{{{x^2}}}} \right]$

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