Implicit differentiation: Survey of methods

Consider an implicit function y = y(x) given by an equation

F(x,y) = c

on a neighborhood of some point. We find its derivative by differentiating both sides of the defining equation, keeping in mind that y is a function of x, and then solving for y′.

More differentiation yields higher order derivatives.

Example: We proved in Solved Problems in Functions - Solved Problems - Implicit and parametric functions that the equation

y³ − 3xy = 1

defines a function on a neighborhood of the point (0,1). Find y′(0).

Solution: We differentiate the equation:

[y³]′ − [3xy]′ = [1]′
3y²[y]′ − (3[x]′y + 3x[y]′) = 0
3y²y′ − 3y − 3xy′ = 0.

Solving for the derivative and then substituting x = 0, y = 1 we get

Remark: The tangent vector to an implicit curve   F(x,y) = C   can be found as   v = (dF/dy,−dF/dx).

Derivative of parametric functions
Back to Methods Survey - Implicit and parametric functions