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* = *f**y*), then with ≠

### Example 1

If *f* (*x*)=*x*^{3} + 2*x* – 10, find (*f*^{-1})'(*x* ).

Step 1: Check if (*f*^{-1})'(*x*) exists. *f* '(*x*)=3*x*^{2} + 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* =*x*^{3} + 2*x* – 10.

Step 3: Interchange *x* and *y* to obtain the inverse function *x* = *y*^{3} +2*y* – 10.

Step 4: Differentiate with respect to *y*:

Step 5: Apply formula

### Example 2

Example 1 could have been done by using implicit differentiation.

Step 1: Le y = *f* (*x*), and thus *y* =*x*^{3} + 2*x* – 10.

Step 2: Interchange *x* and *y* to obtain the inverse function *x* = *y*^{3} +2*y* – 10.

Step 3: Differentiate each term implicitly with respect to *x*.

Step 4: Solve for

### Example 3

If *f* (*x*)= 2*x*^{5} + *x*^{3} + 1, find (a) *f* (1) and *f*'(1) and (b) (*f*^{-1})(4) and (*f*^{-1})'(4). Enter *y*1=2*x*^{5} + *x*^{3} + 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*)=2*x*^{5}+*x*^{3}+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*)=10*x*^{4} + 3*x*^{2}

*f* '(1)=10(1)^{4} +3(1)^{2} =13

### Example 4

If *f*(*x*)=5*x*^{3} + *x* +8, find (*f*^{-1}) '(8).

Enter *y* 1=5*x*^{3} + *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)=15x^{2}+1

f'(0)=1

Practice problems for these concepts can be found at:

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