Derivatives of Inverse Functions for AP Calculus
Practice problems for these concepts can be found at: Differentiation Practice Problems for AP Calculus
Let f be a one-to-one differentiable function with inverse function f-1. If f' (f-1 (a)) ≠0, then the inverse function f-1 is differentiable at a and (f-1)'(a)= . (See Figure 6.5-1.)
If y = f-1 (x) so that x = fy), then with ≠
If f (x)=x3 + 2x – 10, find (f-1)'(x ).
Step 1: Check if (f-1)'(x) exists. f '(x)=3x2 + 2 and f '_(x) > 0 for all real values of x. Thus f(x) is strictly increasing which implies that f (x) is 1 – 1. Therefore, (f-1)'(x) exists.
Step 2: Let y = f (x) and thus y =x3 + 2x – 10.
Step 3: Interchange x and y to obtain the inverse function x = y3 +2y – 10.
Step 4: Differentiate with respect to y:
Step 5: Apply formula
Example 1 could have been done by using implicit differentiation.
Step 1: Le y = f (x), and thus y =x3 + 2x – 10.
Step 2: Interchange x and y to obtain the inverse function x = y3 +2y – 10.
Step 3: Differentiate each term implicitly with respect to x.
Step 4: Solve for
If f (x)= 2x5 + x3 + 1, find (a) f (1) and f'(1) and (b) (f-1)(4) and (f-1)'(4). Enter y1=2x5 + x3 + 1. Since y 1 is strictly increasing, thus f (x) has an inverse.
- f (1)=2(1)5 + (1)3 +1=4
- Since f(1)=4 implies the point (1, 4) is on the curve f(x)=2x5 +x3 +1, therefore, the point (4, 1) (which is the reflection of (1, 4) on y =x) is on the curve (f-1)(x). Thus, (f-1)(4)=1.
f '(x)=10x4 + 3x2
f '(1)=10(1)4 +3(1)2 =13
If f(x)=5x3 + x +8, find (f-1) '(8).
Enter y 1=5x3 + x +8. Since y 1 is strictly increasing near x =8, f(x) has an inverse near x =8.
Note that f(0)=5(0)3 +0+8=8 which implies the point (0, 8) is on the curve of f(x ). Thus, the point (8, 0) is on the curve of (f-1)(x).
f '(x)=15x2 +1
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