Question

Find the derivative of x^x​

Answer 1

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

Given function is

$\rm :\longmapsto\: {x}^{x}$

Let assume that

$\rm :\longmapsto\: y = {x}^{x}$

Taking log on both sides, we get

$\rm :\longmapsto\: logy =log {x}^{x}$

can be rewritten as

$\rm :\longmapsto\: logy =x \: logx$

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

$\rm :\longmapsto\: \dfrac{d}{dx}logy =\dfrac{d}{dx}x \: logx \:$

We know,

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

and

$\boxed{ \tt{ \: \dfrac{d}{dx}uv \: = \: v\dfrac{d}{dx}u \: + \: u\dfrac{d}{dx}v \: }}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{1}{y}\dfrac{dy}{dx} = x \dfrac{d}{dx}logx \: + \: logx \dfrac{d}{dx}x$

$\rm :\longmapsto\:\dfrac{1}{y}\dfrac{dy}{dx} = x \times \dfrac{1}{x} \: + \: logx \times 1$

$\rm :\longmapsto\:\dfrac{1}{y}\dfrac{dy}{dx} = 1 \: + \: logx$

$\bf\implies \:\dfrac{dy}{dx} = y(1 + logx)$

$\bf\implies \:\dfrac{dy}{dx} = {x}^{x} (1 + logx)$

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Short Cut Trick :-

$\boxed{ \tt{ \: \dfrac{d}{dx} {f(x)}^{g(x)} = {f(x)}^{g(x)}\bigg[\dfrac{g(x)}{f(x)}f'(x) + log[f(x)]g'(x) \bigg]}}$

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Additional Information :-

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