Question

e^x-y=x/y find dy/dx​

Answer 1

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

Given expression is

$\rm :\longmapsto\: {e}^{x - y} = \dfrac{x}{y}$

On taking log on both sides, we get

$\rm :\longmapsto\: log{e}^{x - y} =log \dfrac{x}{y}$

$\boxed{\tt{ {log \: x}^{y} = y \: logx \: }} \: \: and \: \: \boxed{\tt{ log \frac{x}{y} = logx - logy \: }}$

So, using this, we get

$\rm :\longmapsto\:(x - y)loge = logx \: - \: logy$

$\rm :\longmapsto\:(x - y) \times 1 = logx \: - \: logy$

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

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

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

$\rm :\longmapsto\:\dfrac{d}{dx}\bigg[logy- y\bigg]= \dfrac{d}{dx}\bigg[logx \: - \: x\bigg]$

We know,

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

So, using this, we get

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

$\rm :\longmapsto\:\dfrac{dy}{dx}\bigg[\dfrac{1}{y} - 1\bigg] = \dfrac{1}{x} - 1$

$\rm :\longmapsto\:\dfrac{dy}{dx}\bigg[\dfrac{1 - y}{y}\bigg] = \dfrac{1 - x}{x}$

$\rm \implies\:\boxed{\tt{ \: \: \dfrac{dy}{dx} \: = \: \frac{y(1 - x)}{x(1 - y)} \: \: }}$

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More to know :-

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