`x e^y = x - y` Find `(dy/dx)` by implicit differentiation.

Textbook Question

Chapter 3, 3.5 - Problem 10 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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hkj1385 | (Level 1) Assistant Educator

Posted on

Note:- 1) If y = e^x ; then dy/dx = e^x

2) If y = x^n ; then dy/dx = n*x^(n-1) ; where 'n' = real number

3) If y = u*v ; where both u & v are functions of 'x' ; then 

dy/dx = u*(dv/dx) + v*(du/dx)

Now, the given function is :-

x*(e^y) = x - y

Differentiating both sides w.r.t 'x' we get

x*(e^y)*(dy/dx) + (e^y) = 1 + (dy/dx)

or, [x*(e^y) - 1]*(dy/dx) = [1 - (e^y)]

or, dy/dx = [1 - (e^y)]/[x*(e^y) - 1]

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balajia | College Teacher | (Level 1) eNoter

Posted on

`xe^y=x-y`

Differentiating both sides with respect to x.

`e^y+xe^y(dy/dx)=1-dy/dx`

`(1+xe^y)(dy/dx)=1-e^y`

`dy/dx=(1-e^y)/(1+xe^y) `

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