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Calculus AB · Unit 3 · 3.3
AP Calculus · Differentiation: Composite, Implicit, and Inverse Functions
Differentiating Inverse Functions
Reflection reciprocates slopes — the only trick is finding which point to reflect from.
key idea$\left(f^{-1}\right)'(a)=\frac{1}{f'\bigl(f^{-1}(a)\bigr)}$
The rule
An inverse function reflects the graph across $y=x$, which turns a slope of $m$ into a slope of $\tfrac1m$. The only difficulty is where to evaluate $f'$ — not at $a$, but at the point that $f$ maps to $a$.
| You want | Evaluate $f'$ at | Because |
|---|
| $\left(f^{-1}\right)'(a)$ | $f^{-1}(a)$ | the reflected point, not $a$ itself |
| $\left(f^{-1}\right)'(5)$ where $f(2)=5$ | $f'(2)$ | $f^{-1}(5)=2$ |
| a slope of $3$ on $f$ | — | gives a slope of $\tfrac13$ on $f^{-1}$ |
| $f'\left(f^{-1}(a)\right)=0$ | — | $f^{-1}$ has a vertical tangent there |
Almost every error is evaluating $f'$ at $a$ instead of at $f^{-1}(a)$. Find the input that $f$ sends to $a$ first.
what the reflection does
- Reflection swaps $x$ and $y$: if $(b,a)$ is on $f$, then $(a,b)$ is on $f^{-1}$.
- Slopes reciprocate under reflection — $m$ becomes $\tfrac1m$.
- $f$ must be one-to-one on the interval for $f^{-1}$ to exist.
- You never need a formula for $f^{-1}$ — only the point it maps from.
Pattern recognition
Given a formulasolve $f(x)=a$ to find $f^{-1}(a)$, then differentiate
Given a tableread the row where the output is $a$
Given a graphreflect across $y=x$ and read the reciprocal slope
Worked examples
EX 1
The standard question
$f(x)=x^{3}+2x+1$For $f(x)=x^{3}+2x+1$, find $\left(f^{-1}\right)'(4)$.
1
Find which input gives $4$
Solve $x^{3}+2x+1=4$, i.e. $x^{3}+2x-3=0$. Testing $x=1$: $1+2-3=0$. ✓ So $f(1)=4$ and $f^{-1}(4)=1$.
2
Differentiate $f$
$f'(x)=3x^{2}+2$.
3
Evaluate at the right point
$f'(1)=3+2=5$ — at $x=1$, not at $x=4$.
4
Reciprocate
$\left(f^{-1}\right)'(4)=\dfrac{1}{5}$.
5
Why no formula was needed
Inverting $x^{3}+2x+1$ algebraically is impossible in closed form. The rule sidesteps that entirely.
EX 2
From a table
structuralGiven $f(3)=7$ and $f'(3)=\tfrac25$, find $\left(f^{-1}\right)'(7)$.
1
Identify the reflected point
$f(3)=7$ means $f^{-1}(7)=3$.
2
Write the rule
$\left(f^{-1}\right)'(7)=\dfrac{1}{f'\left(f^{-1}(7)\right)}=\dfrac{1}{f'(3)}$.
3
Substitute
$=\dfrac{1}{2/5}$.
5
Check plausibility
A shallow slope of $\tfrac25$ on $f$ becomes a steep $\tfrac52$ on $f^{-1}$ — reflection trades shallow for steep.
EX 3
Verify with a known pair
$f(x)=e^{x}$Use the rule on $f(x)=e^{x}$ at $a=1$, and check against the known derivative of $\ln x$.
1
Find $f^{-1}(1)$
$e^{x}=1$ gives $x=0$, so $f^{-1}(1)=0$.
2
Differentiate and evaluate
$f'(x)=e^{x}$, so $f'(0)=1$.
3
Apply the rule
$\left(f^{-1}\right)'(1)=\dfrac{1}{1}=1$.
4
Check independently
$f^{-1}(x)=\ln x$, whose derivative is $\tfrac1x$; at $x=1$ that is $1$. ✓
5
What this confirms
$\dfrac{d}{dx}[\ln x]=\tfrac1x$ is this rule applied to $e^{x}$ — the two facts are the same fact.
EX 4
When the rule fails
$f'=0$$g$ is one-to-one with $g(4)=6$ and $g'(4)=0$. What can you say about $\left(g^{-1}\right)'(6)$?
1
Set up the rule
$\left(g^{-1}\right)'(6)=\dfrac{1}{g'\left(g^{-1}(6)\right)}=\dfrac{1}{g'(4)}$.
2
Substitute
$=\dfrac{1}{0}$ — undefined.
3
Interpret geometrically
A horizontal tangent on $g$ reflects to a vertical tangent on $g^{-1}$, and a vertical tangent has no slope.
4
State the conclusion
$\left(g^{-1}\right)'(6)$ does not exist.
5
Note that $g^{-1}$ still exists
The inverse function is perfectly well defined at $6$; only its derivative fails there.
Watch out
Evaluate $f'$ at $f^{-1}(a)$, never at $a$
This is the error the topic exists to test. In Example 1 you need $f'(1)$, not $f'(4)$.
$\left(f^{-1}\right)'\neq\dfrac{1}{f'(x)}$ as functions
The reciprocal is taken after substituting the reflected input. Reciprocating $f'$ as a function and then evaluating at $a$ gives the wrong answer.
$f^{-1}$ is not $\dfrac{1}{f}$
The superscript $-1$ denotes the inverse function, not a reciprocal. $\sin^{-1}x$ is arcsine, not $\csc x$.
A zero derivative kills the inverse's derivative
$f'\left(f^{-1}(a)\right)=0$ means a vertical tangent on the inverse, so $\left(f^{-1}\right)'(a)$ does not exist.