Question

xy=e^x-y. Find dy/dx.​

Answer 1

Given :

$xy = {e}^{x} - y$

To find :

$\dfrac{dy}{dx}$

Solution :

$\implies \: xy = {e}^{x} - y$

differentiating both sides :-

$\implies \: \dfrac{d(xy)}{dx} = \dfrac{d({e}^{x} - y)}{dx}$

We will differentiate LHS using product rule.

$\implies \: y \dfrac{dx}{dx} + x\dfrac{dy}{dx} = \dfrac{d( {e}^{x}) }{dx} + \dfrac{dy}{dx}$

$\implies \: y + x\dfrac{dy}{dx} = \dfrac{d( {e}^{x}) }{dx} + \dfrac{dy}{dx}$

$\implies \: y + x\dfrac{dy}{dx} = {e}^{x} + \dfrac{dy}{dx}$

$\implies \: x\dfrac{dy}{dx} - \dfrac{dy}{dx} = {e}^{x} - y$

$\implies \: (x - 1) \dfrac{dy}{dx}= ({e}^{x} - y)$

$\implies \: \dfrac{dy}{dx}= ({e}^{x} - y) \times \dfrac{1}{(x - 1)}$

$\implies \: \dfrac{dy}{dx}= \dfrac{ {e}^{x} - y}{x - 1}$

Answer :

$\dfrac{dy}{dx}= \dfrac{ {e}^{x} - y}{x - 1}$

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$\large\bf\blue{Additional-Information}$

d(sinx)/dx = cosx

d(cos x)/dx = -sin x

d(cosec x)/dx = -cot x cosec x

d(tan x)/dx = sec²x

d(sec x)/dx = secx tanx

d(cot x)/dx = - cosec² x

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