# Use Implicit Differentiation to find dy/dx and d^2y/dx^2. x^(2/3)+y^(2/3)=1 Thanks so much for your help!!

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Given `x^(2/3)+y^(2/3)=1`

Taking the first derivative implicitly we get:

`2/3x^(-1/3)+2/3y^(-1/3)(dy)/(dx)=0`

So

`2/3 y^(-1/3)(dy)/(dx)=-2/3x^(-1/3)`

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`(dy)/(dx)=-x^(-1/3)y^(1/3)`

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Then

`(d^2y)/(dx^2)=-[x^(-1/3)(1/3)y^(-2/3)(dy)/(dx)-1/3x^(-4/3)y^(1/3)]` substituting for `(dy)/(dx)`

`=-[x^(-1/3)(1/3)y^(-2/3)(-x^(-1/3)y^(1/3))-1/3x^(-4/3)y^(1/3)]`

`=-[-1/3x^(-2/3)y^(-1/3)-1/3x^(-4/3)y^(1/3)]`

`=1/3x^(-4/3)y^(-1/3)[x^(2/3)+y^(2/3)]` But `x^(2/3)+y^(2/3)=1` so

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`(d^2y)/(dx^2)=1/3x^(-4/3)y^(-1/3)`

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