Question

log (xy)= x+y find dy/dx​

Answer 1

Step-by-step explanation:

➡given here .

↪log(xy)= (x+y).

▶Differentiate w.r. to x .

↪d/dx ( logxy )= d/dx(x+y).

↪1/xy (y+xdy/dx)= 1+dy/dx.

↪1/x - 1. = ( -1/y +1 ) dy/dx .

↪dy/dx = y(1-x)/(y-1)x

Answer 2

$\huge \mathfrak{solution}$

Given:

$log(xy) = x + y$

Differentiate both sides w.r.t X

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

Formula:

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

$\boxed{ \green{ \bold{\frac{d}{dx} (u.v) = u. \frac{d}{d \: x} (v) + v. \frac{d}{dx} (u)}}}$