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:



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





`(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)`



`=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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