Question

$\boxed {\red{\mathtt{find \: \: \frac{dy}{dx} \: \: if \: \: {x}^{y} \cdot {y}^{x} = {x}^{x}} }}$
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Answer 1

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

Given function is

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

Taking log on both sides, we get

$\rm :\longmapsto\: log[{x}^{y}. {y}^{x}]= log {x}^{x}$

We know,

$\boxed{\tt{ logxy = logx + logy}} \\ \\ and \\ \\\boxed{\tt{ \: \: log {x}^{y} = ylogx \: \: }}$

So, using these, we get

$\rm :\longmapsto\:ylogx + xlogy = xlogx$

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

$\rm :\longmapsto\:\dfrac{d}{dx}\bigg[ylogx + xlogy\bigg] = \dfrac{d}{dx}xlogx$

$\rm :\longmapsto\:\dfrac{d}{dx}(ylogx) + \dfrac{d}{dx}(xlogy) = \dfrac{d}{dx}xlogx$

We know,

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

So, on applying this, we have

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

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

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

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

$\rm :\longmapsto\: \bigg[logx + \dfrac{x}{y}\bigg]\dfrac{dy}{dx} = \dfrac{x + xlogx - y - xlogy}{x}$

$\rm :\longmapsto\: \bigg[\dfrac{ylogx + x}{y}\bigg]\dfrac{dy}{dx} = \dfrac{x + xlogx - y - xlogy}{x}$

$\rm :\longmapsto\: \dfrac{dy}{dx} = \dfrac{y[x + xlogx - y - xlogy]}{x[ylogx + x]}$

Hence,

$\rm\implies \:\boxed{\tt{ \dfrac{dy}{dx} = \dfrac{y[x + xlogx - y - xlogy]}{x[ylogx + x]} }}$

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MORE TO KNOW

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