AP Calculus AB/BC · Unit 3 · Topic 3.3
Differentiating Inverse Functions
Use reciprocal slopes at corresponding input-output points.
1. Topic Focus
Differentiate nested, implicit, inverse, and higher-order relationships by choosing procedures that match the function structure.
This topic: Use reciprocal slopes at corresponding input-output points.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
If f(2)=5 and f′(2)=3, then (f⁻¹)′(5)=1/3.
Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.
5. Concept Development
An inverse function reverses a one-to-one input-output relationship. If \(f(a)=b\), then
The point \((a,b)\) on \(y=f(x)\) therefore corresponds to \((b,a)\) on \(y=f^{-1}(x)\). Their graphs are reflections across \(y=x\), so a nonzero tangent slope is reversed from rise-over-run to run-over-rise.
Equivalently, at a general input \(x\),
Why the formula works. The inverse identities give \(f(f^{-1}(x))=x\). Differentiating both sides with the chain rule yields
and division by the original derivative produces the inverse-derivative formula.
| Original function \(f\) | Inverse function \(f^{-1}\) |
|---|---|
| Point \((a,b)\) | Point \((b,a)\) |
| Input \(a\), output \(b\) | Input \(b\), output \(a\) |
| Slope \(f'(a)=m\ne0\) | Slope \((f^{-1})'(b)=1/m\) |
| Domain of \(f\) | Range of \(f^{-1}\) |
| Range of \(f\) | Domain of \(f^{-1}\) |
AP evaluation routine. To find \((f^{-1})'(c)\):
- Locate the original input \(a\) whose output is \(c\); that is, solve or read \(f(a)=c\).
- Evaluate the original derivative at that input: \(f'(a)\).
- Take the reciprocal: \((f^{-1})'(c)=1/f'(a)\).
The most common mistake is using \(1/f'(c)\). The denominator is evaluated at \(a=f^{-1}(c)\), not automatically at the inverse function's input \(c\).
Conditions and local meaning. A global inverse requires \(f\) to be one-to-one on its stated domain. A domain restriction can make a function such as \(x^2\) invertible. At the corresponding point, the standard derivative formula requires \(f'(a)\ne0\). If \(f'(a)=0\), the inverse may have a vertical tangent or may fail to be differentiable, so writing \(1/0\) is not a finite derivative.
Composite inverse functions. If \(H(x)=f^{-1}(g(x))\), the outer inverse rule and inner chain rule combine:
First follow \(x\) through \(g\), then through \(f^{-1}\); the derivative of \(f\) is evaluated at the resulting original-function input.
Units reverse too. If \(f'(a)\) is measured in output units per input unit, then \((f^{-1})'(b)\) is measured in input units per output unit. This reciprocal-unit check often catches an inverted or incorrectly located derivative value.
6. Detailed Worked Example and Error Check
Example 1: Direct values. Suppose \(f(2)=5\) and \(f'(2)=3\). Because \(f^{-1}(5)=2\),
The incorrect expression \(1/f'(5)\) ignores the input-output reversal.
Example 2: No explicit inverse formula needed. Let \(f(x)=x^3+x\). Since \(f'(x)=3x^2+1>0\), \(f\) is one-to-one. Also \(f(2)=10\), so
Solving the cubic for a complete formula for \(f^{-1}\) would add work without helping this point evaluation.
Example 3: Read from a table.
| \(x\) | \(f(x)\) | \(f'(x)\) |
|---|---|---|
| \(-1\) | \(6\) | \(4\) |
| \(1\) | \(2\) | \(-3\) |
| \(4\) | \(-3\) | \(-5\) |
To find \((f^{-1})'(-3)\), locate \(-3\) in the \(f(x)\) column. It occurs at \(x=4\), so
Example 4: An inverse with no elementary formula. Let \(f(x)=x+e^x\), and let \(g=f^{-1}\). Since \(f(0)=1\), \(g(1)=0\). Therefore
Example 5: Inverse function inside a composition. Suppose \(H(x)=f^{-1}(x^2+1)\), \(f(3)=5\), and \(f'(3)=-4\). At \(x=2\), the inner output is \(5\), so \(f^{-1}(5)=3\). Thus
Example 6: Tangent line to an inverse. Let \(f(x)=x^3+2x\). The original point \((1,3)\) becomes \((3,1)\) on \(f^{-1}\). Since \(f'(1)=5\), the inverse slope is \(1/5\), and the tangent line is
Example 7: Interpret reciprocal units. Suppose \(T(t)\) is temperature in degrees Celsius at time \(t\) seconds and \(T'(4)=2.5\) degrees per second. If \(T\) is locally one-to-one, then
seconds per degree Celsius. Near that temperature, an additional degree corresponds to about \(0.4\) additional second.
AP error check. Do not confuse \(f^{-1}(x)\) with \(1/f(x)\), evaluate \(f'\) at the inverse input instead of the corresponding original input, take the reciprocal of a function value, ignore a domain restriction needed for invertibility, or claim a finite inverse derivative when the original slope is zero.
7. AP Reasoning Routine
Mark inner and outer functions, track every derivative factor, solve algebraically for the requested derivative, and verify the result's domain.
- Identify the representation and requested quantity.
- State the rule or theorem and verify its conditions.
- Keep exact values until the final requested approximation.
- Interpret sign, units, interval, and context.
For each problem, identify the corresponding point on the original function before taking a reciprocal.
(a) If \(f(-2)=7\) and \(f'(-2)=-6\), find \((f^{-1})'(7)\).
(b) Let \(f(x)=x^3+4x\). Find \((f^{-1})'(5)\).
(c) A table gives \(f(0)=4\), \(f'(0)=2\), \(f(3)=0\), and \(f'(3)=-7\). Find \((f^{-1})'(0)\).
(d) Let \(f(x)=x^2\) on \([0,\infty)\). Use the inverse-derivative theorem to find \((f^{-1})'(9)\).
(e) Suppose \(g=f^{-1}\), \(f(2)=-1\), and \(f'(2)=4\). Find the tangent line to \(g\) at input \(-1\).
(f) If \(H(x)=f^{-1}(3x-1)\), \(f(4)=5\), and \(f'(4)=6\), find \(H'(2)\).
(g) A distance function \(D(t)\), measured in meters, satisfies \(D'(8)=3\) meters per second. Interpret \((D^{-1})'(D(8))\).
(h) Explain what can go wrong with the formula for the inverse of \(f(x)=x^3\) at \(x=0\).
Check the solution
(a) Since \(f^{-1}(7)=-2\), \((f^{-1})'(7)=1/f'(-2)=-1/6\).
(b) \(f(1)=5\) and \(f'(x)=3x^2+4\), so \((f^{-1})'(5)=1/f'(1)=1/7\).
(c) Because \(f(3)=0\), \((f^{-1})'(0)=1/f'(3)=-1/7\).
(d) \(f^{-1}(9)=3\) and \(f'(3)=6\), so \((f^{-1})'(9)=1/6\), agreeing with the derivative of \(\sqrt{x}\) at 9.
(e) The inverse point is \((-1,2)\), and its slope is \(1/4\). The tangent line is \(y-2=\frac14(x+1)\).
(f) At \(x=2\), \(3x-1=5\) and \(f^{-1}(5)=4\). Therefore \(H'(2)=3/f'(4)=1/2\).
(g) Near the distance \(D(8)\), elapsed time changes at \(1/3\) second per additional meter.
(h) Although \(f(x)=x^3\) is one-to-one, \(f'(0)=0\), so the reciprocal formula does not give a finite value. Its inverse \(x^{1/3}\) has a vertical tangent and is not differentiable at 0 in the ordinary finite-slope sense.