`x^3 + y^3 = 1` Find `(dy/dx)` by implicit differentiation.

Textbook Question

Chapter 3, 3.5 - Problem 5 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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hkj1385's profile pic

hkj1385 | (Level 1) Assistant Educator

Posted on

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

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

Now, 

(x^3) + (y^3) = 1

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

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

or, dy/dx = -[x^2]/[y^2]

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balajia | College Teacher | (Level 1) eNoter

Posted on

`x^3+y^3=1`

differentiating with respect to x.

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

`dy/dx=-x^2/y^2`

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