Implicit Differentiation

Consider the equation (y² + x² - 1)(x+y)(x-y) = 0. The graph of this equation is the graph of the circle of radius 1 centered at the origin, together with the graphs of the lines which bisect the quadrants.

This is clearly not the graph of a function, yet at many points on the graph there a re clearly going to be tangent lines. The method by which the slopes of these tangent lines can be found is called implicit differentiation. The idea is that if we look in the vicinity of points where there are tangent lines, frequently we see the graph of a function, so it is reasonable to believe that at such points y is a differentiable function of x. Hence the chain rule, the product rule, and other rules for differentiation may be applied to find y' in terms of the coordinates of the point. Here is a simple example:

x² + 3xy + 4y² + 2x + 3y = 13

It can be shown (without calculus!!) that the graph of this equation is an ellipse, and that (1,1) is a point on this graph. What is an equation of the tangent line to this graph at (1,1)? Presuming that y is a function of x near (1,1) we have

2x + 3(xy' + (1)y) + 8yy' + 2 + 3y' = 0

Therefore, at (1,1),

2(1) + 3((1)y' + (1)(1)) + 8(1)y' + 2 + 3y' = 0.

Solving for y' gives y' = -7/14 = -1/2, so the tangent line has equation y-1 = -(x-1)/2.

Related problems might be to find the points where y' = 0 and where y' is undefined. To solve the former problem, note that since

2x + 3(xy' + (1)y) + 8yy' + 2 + 3y' = 0

we can expect that

$$y' = -\frac{2x + 3y + 2}{3x + 8y + 3} \;{\rm if}\; 3x +8y + 3 \neq 0$$

so we are looking for points (x,y) where

$$\begin{array} {rcl} 2x + 3y + 2 & = & 0\\ x^2 + 3xy + 4y^2 + 2x + 3y & = & 13\end{array}$$

For the latter problem, we are looking for points where

$$\begin{array} {rcl} 3x + 8y + 3 & = & 0\\ x^2 + 3xy + 4y^2 + 2x + 3y & = & 13\end{array}$$

Either set of equations may be solved by eliminating y and solving a quadratic equation for x.