`x^2 + xy - y^2 = 4` Find `(dy/dx)` by implicit differentiation.

Textbook Question

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

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

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

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

3) If y = k ; where k = constant ; then dy/dx = 0

Now, the given function is :- 

(x^2) + xy - (y^2) = 4

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

2x + x*(dy/dx) + y - 2y*(dy/dx) = 0

or, (2x+y) = (2y-x)*(dy/dx)

or, dy/dx = (2x+y)/(2y-x)

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balajia | College Teacher | (Level 1) eNoter

Posted on

`x^2+xy-y^2=4`

Differentiating with respect to x. We get

`2x+(y+x(dy/dx))-2y(dy/dx)=0`

`(2x+y)+(x-2y)(dy/dx)=0`

`dy/dx=(2y-x)/(2x+y)`

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