Implicit Differentiation
Implicit differentiation is a special case of the chain rule for derivatives. Some of the type of questions that require knowledge of implicit differentiation are described below.
1. A curve in the (x, y) plane is given and the problem may ask to calculate dy/dx, dx/dy. Examples.
2. Another type of problem asks for the line tangent to a given curve in a specific point. In most case, the problems can be solved by treating x as an implicitly defined function of y and differentiating both sides of the equation with respect to x.
3. A similar type of question may ask to find the points in the (x, y) plane where a specific curve has vertical or horizontal tangents. If the horizontal tangents need to be determined, the problems can be solved by treating x as an implicitly defined function of y and differentiating both sides of the equation with respect to x. If the vertical tangents need to be determined, the problems can be solved by treating y as an implicitly defined function of x and differentiating both sides of the equation with respect to y. Examples.
Examples
Question 1:
Consider the curve given by: x3 + xy = 2y2 + y - 5
Show that:
3x2 + xy'+ y = 4yy' + y'
Question 2:
Find the coordinates of the two points on the x2 - 2x + 4y2 + 16y + 1 = 0 closed curve where the line tangent to the curve is vertical.
Treating y as an implicitly defined function of x and differentiating both sides of the equation with respect to y, we obtain:
2xx' - 2x' + 8y + 16 = 0
x'(2x - 2) = -8(y + 2)
x' = -4(y + 2)/(x - 1) and x' should be 0 if the line tangent is vertical.
y = -2
At the points with vertical tangents, x2 - 2x + 16 - 32 + 1 = 0
x2 - 2x - 15 = 0
The solutions are x1 = 5 and x2 = -3.
The coordinates of the two points are (5, -2) and (-3, -2).
The official AP Calculus webpage