Calculus AB and Calculus BC

CHAPTER 3 Differentiation

H. DERIVATIVE OF THE INVERSE OF A FUNCTION

Suppose f and g are inverse functions. What is the relationship between their derivatives? Recall that the graphs of inverse functions are the reflections of each other in the line y = x, and that at corresponding points their x- and y-coordinates are interchanged.

Figure N3–8 shows a function f passing through point (a,b) and the line tangent to f at that point. The slope of the curve there, f (a), is represented by the ratio of the legs of the triangle, Image When this figure is reflected across the line y = x, we obtain the graph of f −1, passing through point (b,a), with the horizontal and vertical sides of the slope triangle interchanged. Note that the slope of the line tangent to the graph of f −1 at x = b is represented by Image the reciprocal of the slope of f at x = a. We have, therefore,

Image

Image

FIGURE N3–8

Simply put, the derivative of the inverse of a function at a point is the reciprocal of the derivative of the function at the corresponding point.

EXAMPLE 32

If f (3) = 8 and f (3) = 5, what do we know about f −1?

SOLUTION: Since f passes through the point (3,8), f −1 must pass through the point (8,3). Furthermore, since the graph of f has slope 5 at (3,8), the graph of f −1 must have slope Image at (8,3).

EXAMPLE 33

A function f and its derivative take on the values shown in the table. If g is the inverse of f, find g (6).

SOLUTION: To find the slope of g at the point where x = 6, we must look at the point on f where y = 6, namely, (2,6). Since f (2) = Image g (6) = 3.

x

f (x)

f (x)

2

6

Image

6

8

Image

EXAMPLE 34

Let y = f (x) = x3 + x − 2, and let g be the inverse function. Evaluate g (0).

SOLUTION: Since Image To find x when y = 0, we must solve the equation x3 + x − 2 = 0. Note by inspection that x = 1, so

Image

EXAMPLE 35

Where is the tangent to the curve 4x2 + 9y2 = 36 vertical?

SOLUTION: We differentiate the equation implicitly to get Image so Image Since the tangent line to a curve is vertical when Image we conclude that Image must equal zero; that is, y must equal zero. When we substitute y = 0 in the original equation, we get x = ±3. The points (±3,0) are the ends of the major axis of the ellipse, where the tangents are indeed vertical.