Question

if xy=log(xy)=log(xy) then show that dy/dx= -y/x​

Answer 1

Given:

$\star \: \: \bold{ xy = log(xy) }$

To show :

$\star \: \bold{ \frac{dy}{dx} = - \frac{ y}{x} }$

Solution:

$\implies \: xy = log(xy)$

Differentiate w.r.t both sides :

$\implies \: \frac{d}{dx} (xy) = \frac{d}{dx} ( log(xy) ) \\ \\ \implies \: x \: \frac{d}{dx} (y) + y \: \frac{d}{dx} (x) = \frac{1}{ (xy) } \times \frac{d}{dx} (xy) \\ \\ \implies \: x \frac{dy}{dx} + y(1) = \frac{1}{xy} (x \frac{dy}{dx} + y) \\ \\ \implies \: x \frac{dy}{dx} + y = \frac{ \cancel{x}}{ \cancel{x}y} \frac{dy}{dx} + \frac{ \cancel{y}}{x \cancel{y}} \\ \\ \implies \: x \frac{dy}{dx} + y = \frac{1}{y} \frac{dy}{dx} + \frac{1}{x} \\ \\ \implies \: x \frac{dy}{dx} - \frac{1}{y} \frac{dy}{dx} = \frac{1}{x} - y \\ \\ \implies \: \frac{dy}{dx} (x - \frac{1}{y} ) = \frac{1 - xy}{x} \\ \\ \implies \: \frac{dy}{dx} ( \frac{xy - 1}{y} ) = - ( \frac{xy - 1}{x} ) \\ \\ \implies \: \frac{dy}{dx} ( \frac{ \cancel {xy - 1}}{y} ) = - ( \frac{ \cancel{xy - 1}}{x} ) \\ \\ \implies \: \frac{dy}{dx} = - \frac{y}{x}$

Hence proved .

Formula used :

$\star \boxed{ \bold{ \red{ \frac{d}{dx} (u.v) = u \: \frac{dv}{dx} + v \: \frac{du}{dx} }}}$

$\star \boxed{ \green{\bold{ \frac{d}{dx} log(x) = \frac{1}{x} }}}$