Derivative Of An Implicit Function

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Derivative of An Implicit Function

Suppose that two variable x and y are related by a given implicit relation f (x,y) = 0. In general, this defines y as a multi-valued function of x. However, the simplest way of dealing with such a function is to introduce a third variable z by

z = f (x,y).

Of course, z = 0 for all x and y related by f (x,y) = 0. This the given implicit function can be studied by relating those values of x and y which make z equal to zero in the explicit function z = f (x,y).

As x and y vary in any way, independently or not, the complete differential of z is given by

dz = fx dx + fy dy.

If (x,y) are values making z = 0 and if dx and dy  are variations from these values keeping z = 0, then dz = 0. That is,

fx dx + fy dy = 0.

Hence

dy/dx = - fx / fy

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