Question

Find $\dfrac{dy}{dx}$ for :
$e^{x-y}=log(\dfrac{x}{y})$​

Answer 1

Given : $\mathrm{e^{x-y}=log\frac{x}{y}}$

$\implies \mathrm{\frac{e^{x}}{e^{y}}=logx-logy}$

Differentiating both sides with respect to x, we get

$\mathrm{\frac{d}{dx}\frac{e^{x}}{e^{y}}=\frac{d}{dx}(logx-logy)}$

$\to \small{\mathrm{\frac{e^{y}\frac{d}{dx}(e^{x})-e^{x}\frac{d}{dx}(e^{y})}{(e^{y})^{2}}=\frac{d}{dx}(logx)-\frac{d}{dx}(logy)}}$

$\to \mathrm{\frac{e^{y}e^{x}-e^{x}e^{y}\frac{dy}{dx}}{(e^{y})^{2}}=\frac{1}{x}-\frac{1}{y}\frac{dy}{dx}}$

$\to \mathrm{\frac{e^{x}e^{y}\left(1-\frac{dy}{dx}\right)}{(e^{y})^{2}}=\frac{1}{x}-\frac{1}{y}\frac{dy}{dx}}$

$\to \mathrm{\frac{e^{x}\left(1-\frac{dy}{dx}\right)}{e^{y}}=\frac{1}{x}-\frac{1}{y}\frac{dy}{dx}}$

$\to \mathrm{\frac{e^{x}}{e^{y}}-\frac{e^{x}}{e^{y}}\frac{dy}{dx}=\frac{1}{x}-\frac{1}{y}\frac{dy}{dx}}$

$\to \mathrm{\frac{1}{y}\frac{dy}{dx}-\frac{e^{x}}{e^{y}}\frac{dy}{dx}=\frac{1}{x}-\frac{e^{x}}{e^{y}}}$

$\to \mathrm{\left(\frac{1}{y}-\frac{e^{x}}{e^{y}}\right)\frac{dy}{dx}=\frac{e^{y}-xe^{x}}{xe^{y}}}$

$\to \mathrm{\frac{e^{y}-ye^{x}}{ye^{y}}\frac{dy}{dx}=\frac{e^{y}-ye^{x}}{xe^{y}}}$

$\to \mathrm{\frac{dy}{dx}=\frac{y(e^{y}-xe^{x})}{x(e^{y}-ye^{x})},\:e^{y}\neq 0}$

Answer 2

Answer:

Step-by-step explanation:

Differentiation:-

First we take log both sides,

We get:-

x-y=log(log(x/y))

Now,we differentiate both sides with respect to x.

Applying chain rule,we get the final answer.

Formulae used:-

log e^x=x

(d/dx)(logx)=1/x

(d/dx)(u/v)={v(du/dx)-u(dv/dx)}/(v*v)