Question

$\boxed\scriptsize{\tt\pink{Differentiate \quad With \quad Respect \quad to\: X.}}$
$\boxed{\tt \red{y=(\sin x)^{x}+\left(\frac{1}{x}\right)^{\cos x}}}$
don't Spam.
answer with full explanation.
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Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:\tt {y=(\sin x)^{x}+\left(\dfrac{1}{x}\right)^{\cos x}}$

Let assume that,

$\rm :\longmapsto\:y = u + v$

So that,

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{du}{dx} + \dfrac{dv}{dx} - - - - (1)$

where,

$\rm :\longmapsto\:u = {(sinx)}^{x} - - - (2)$

and

$\rm :\longmapsto\:v = {\bigg(\dfrac{1}{x} \bigg) }^{cosx} - - - (3)$

Now, Consider

$\rm :\longmapsto\:u = {(sinx)}^{x}$

Taking log on both sides, we get

$\rm :\longmapsto\:logu = log{(sinx)}^{x}$

$\rm :\longmapsto\:logu =x \: log{(sinx)}$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}logu =\dfrac{d}{dx}x \: log{(sinx)}$

We know,

$\boxed{\tt{ \dfrac{d}{dx}logx = \frac{1}{x}}} \\ \\ \boxed{\tt{ \dfrac{d}{dx}uv = \: u\dfrac{d}{dx}v \: + \: v\dfrac{d}{dx}u}}$

So, using this result, we get

$\rm :\longmapsto\:\dfrac{1}{u}\dfrac{du}{dx} = x\dfrac{d}{dx}logsinx + logsinx\dfrac{d}{dx}x$

$\rm :\longmapsto\:\dfrac{1}{u}\dfrac{du}{dx} = x \times \dfrac{1}{sinx}\dfrac{d}{dx}sinx + logsinx \times 1$

$\rm :\longmapsto\:\dfrac{1}{u}\dfrac{du}{dx} = \dfrac{x}{sinx}(cosx) + logsinx$

$\rm :\longmapsto\:\dfrac{du}{dx} = u(x \: cotx + logsinx)$

$\rm\implies \:\boxed{\tt{ \dfrac{du}{dx} = {(sinx)}^{x}(x \: cotx + logsinx)}} - - (4)$

Now, Consider

$\rm :\longmapsto\:v = {\bigg[\dfrac{1}{x} \bigg]}^{cosx}$

can be rewritten as

$\rm :\longmapsto\:v = {x}^{ - cosx}$

Taking log on both sides, we get

$\rm :\longmapsto\:logv = log{(x)}^{ - cosx}$

$\rm :\longmapsto\:logv = - cosx \: log{(x)}$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}logv = -\dfrac{d}{dx} cosx \: log{(x)}$

$\rm :\longmapsto\:\dfrac{1}{v}\dfrac{dv}{dx} = - \bigg(cosx\dfrac{d}{dx}logx + logx\dfrac{d}{dx}cosx\bigg)$

$\rm :\longmapsto\:\dfrac{1}{v}\dfrac{dv}{dx} = - \bigg(cosx \times \dfrac{1}{x} + logx( - sinx)\bigg)$

$\rm :\longmapsto\:\dfrac{dv}{dx} = - v\bigg[\dfrac{cosx}{x} - sinx \: logx\bigg]$

$\rm\implies \:\boxed{\tt{ \dfrac{dv}{dx} = - {(x)}^{ -cosx} \bigg[\dfrac{cosx}{x} - sinx \: logx\bigg]}} - - (5)$

On substituting the values of equation (4) and (5), in equation (1), we get

$\red{\sf :\longmapsto\:\dfrac{dy}{dx} = {(sinx)}^{x}(xcotx + logsinx)- {(x)}^{- cosx} \bigg[\dfrac{cosx}{x} - sinx logx\bigg]}$

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FORMULA USED

$\boxed{\tt{ log {x}^{y} \: = \: y \: logx \: }}$

$\boxed{\tt{ \dfrac{d}{dx}sinx = cosx}}$

$\boxed{\tt{ \dfrac{d}{dx}cosx \: = - \: sinx}}$

$\boxed{\tt{ \dfrac{d}{dx}x = 1}}$

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$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$