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

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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 = 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]