# Find the derivative Find dy/dx if x^2/y^2  = 0, by using implicit differentiation. To differentiate a function f(y) of y with respect to x we use the chain rule as follows:

`d/(dx) f(y) = d/(dy) f(y)*(dy)/(dx) = f'(y)(dy)/(dx)`

Differentiating `x^2/y^2` implicitly with respect to `x` use this result and the multiplication rule of differentiation. So

`d/(dx) (x^2/y^2) = (d/(dx) x^2)/y^2 +(x^2)d/(dy)(y^-2) (dy)/(dx)`

`=...

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To differentiate a function f(y) of y with respect to x we use the chain rule as follows:

`d/(dx) f(y) = d/(dy) f(y)*(dy)/(dx) = f'(y)(dy)/(dx)`

Differentiating `x^2/y^2` implicitly with respect to `x` use this result and the multiplication rule of differentiation. So

`d/(dx) (x^2/y^2) = (d/(dx) x^2)/y^2 +(x^2)d/(dy)(y^-2) (dy)/(dx)`

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

Setting `x^2/y^2 = 0` we have

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

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

`therefore (dy)/(dx) = (2xy^3)/(2x^2y^2) = y/x` answer

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