Find dy/dx : x^3+y^3=18y

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rubeus-hagrid | (Level 1) Adjunct Educator

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x^3 + y^3 = 18 y

3x^2 + 3y^2 dy/dx = 18 dy/dx

3x^2 = 18 dy/dx - 3y^2 dy/dx

3x^2 = dy/dx (18-3y^2)

(3x^2)/(18-3y^2) = dy/dx

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to find dy/dx given that x^3+y^3=18y

x^3 + y^3 = 18*y

Using implicit differentiation gives:

[x^3 + y^3]' = [18*y]'

=> [x^3]' + [y^3]' = [18*y]'

=> 3x^2 + 3y^2*y' = 18*y'

=> y'(18 - 3y^2) = 3x^2

=> y' = 3x^2/(18 - 3y^2)

The required value of dy/dx = 3x^2/(18 - 3y^2)

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