`2x^3 + x^2 y - xy^3 = 2` Find `(dy/dx)` by implicit differentiation.

Textbook Question

Chapter 3, 3.5 - Problem 8 - 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 :-

2*(x^3) + (x^2)*y - x*(y^3) = 2

differentiating w.r.t 'x' on both sides we get,

6*(x^2) + 2xy + (x^2)*(dy/dx) - (y^3) - 3x*(y^2)*(dy/dx) = 0

or, [6(x^2) + 2xy - (y^3)] = [3x*(y^2) - (x^2)]*(dy/dx)

or, dy/dx = [6(x^2) + 2xy - (y^3)]/[3x*(y^2) - (x^2)]

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balajia | College Teacher | (Level 1) eNoter

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`2x^3+x^2y-xy^3=2`

Differentiating with respect to x.we get

`2(3x^2)+(2xy+x^2(dy/dx))-(y^3+x(3y^2)(dy/dx))=0`

`(6x^2+2xy-y^3)+(x^2-3xy^2)(dy/dx)=0`

`dy/dx=(6x^2+2xy-y^3)/(3xy^2-x^2)`

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