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 ycoordinates 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, 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 the reciprocal of the slope of f at x = a. We have, therefore,
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 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) = g ′(6) = 3.
x 
f (x) 
f ′(x) 
2 
6 

6 
8 
EXAMPLE 34
Let y = f (x) = x^{3} + x − 2, and let g be the inverse function. Evaluate g ′(0).
SOLUTION: Since To find x when y = 0, we must solve the equation x^{3} + x − 2 = 0. Note by inspection that x = 1, so
EXAMPLE 35
Where is the tangent to the curve 4x^{2} + 9y^{2} = 36 vertical?
SOLUTION: We differentiate the equation implicitly to get so Since the tangent line to a curve is vertical when we conclude that 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.