Question

​Differentiate With Respect to X.
$\purple{ \boxed{ \tt\pink{\bold{y=x^{x} \cdot \sin x+(\sin x)^{x}}}}}$
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Answer 1

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

Given function is

$\rm :\longmapsto\:y=x^{x} \cdot \sin x+(\sin x)^{x}$

Let assume that

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

So, that

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

where,

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

and

$\rm :\longmapsto\:v = {(sinx)}^{x} - - - - (3)$

Now, Consider

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

Taking log on both sides, we get

$\rm :\longmapsto\:logu = log[{x}^{x} \: sinx]$

$\rm :\longmapsto\:logu = xlogx + logsinx$

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

$\rm :\longmapsto\:\dfrac{d}{dx}logu =\dfrac{d}{dx}[ xlogx + logsinx]$

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

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

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

$\rm :\longmapsto\:\dfrac{1}{u}\dfrac{du}{dx} = 1+ logx + cotx$

$\rm :\longmapsto\:\dfrac{du}{dx} = u\bigg[1+ logx + cotx\bigg]$

$\red{\rm\implies \:\dfrac{du}{dx} = {x}^{x}sinx \bigg[1+ logx + cotx\bigg] - - - (4)}$

Now, Consider

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

On taking log on both sides, we get

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

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

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

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

$\rm :\longmapsto\:\dfrac{1}{v}\dfrac{dv}{dx} = x\dfrac{d}{dx}logsinx + logsinx\dfrac{d}{dx}x$

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

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

$\rm :\longmapsto\: \dfrac{dv}{dx} = v\bigg[x \: cotx + logsinx\bigg]$

$\green{\rm\implies \: \dfrac{dv}{dx} = {(sinx)}^{x} \bigg[x \: cotx + logsinx\bigg] - - - (5)}$

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

$\purple{\rm :\longmapsto\:\dfrac{dy}{dx} = {x}^{x}sinx \bigg[1+ logx + cotx\bigg] + {(sinx)}^{x} \bigg[x \: cotx + logsinx\bigg]}$

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FORMULA USED

$\boxed{\tt{ \dfrac{d}{dx} {x}^{y} = y \: logx}}$

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

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

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

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

Answer 2

Answer:

Step-by-step explanation:

Let $u= x^{x}sinx$ and $v= (sinx)^{x}$

$logu = log(x^{x} sinx) =log(x^{x}) + log(sinx) =xlogx + log(sinx)\\\frac{1}{u}\frac{du}{dx}= 1(logx)+\frac{x}{x} + \frac{1}{sinx} cosx = logx+ 1+ cotx\\ \frac{du}{dx} = (x^{x} sinx)(logx+ 1+ cotx)$

$log v = xlog(sinx)\\\frac{1}{v} \frac{dv}{dx} = 1[log(sinx)] + x(\frac{1}{sinx} cosx) = log(sinx) + xcotx\\\frac{dv}{dx}=(sinx)^{x} (log(sinx) + xcotx)$

$\frac{dy}{dx} =\frac{du}{dx} +\frac{dv}{dx}\\=(x^{x}sinx)(logx+1+cotx) +(sinx)^{x}[log(sinx) + xcotx)$