Calculus I · Unit 2A · lesson

Derivative of an Inverse Function

Concept

Learning objectives

Use the inverse-function derivative theorem numerically and symbolically.

Reciprocal Slopes of Inverse Functions

Explanation

Before the formulas

In Derivative of an Inverse Function, inverse and implicit ideas meet. Swapping input and output swaps horizontal and vertical change, so inverse slopes are reciprocals at corresponding points. Taking logarithms can also reveal hidden structure by turning products into sums and exponents into coefficients.

These methods are strategic transformations, not new definitions of derivative. State the domain assumptions, preserve the original relationship, and substitute back at the end. A clean solution explains why the transformation helps before carrying out the algebra.

**Visual study**
Reflecting a graph across y=x swaps horizontal and vertical changes. A tangent with slope m reflects to one with slope 1/m.

Explanation

Inverse functions swap inputs and outputs, so slopes become reciprocals

The graph of an inverse function is the reflection of the original graph across $y=x$ . A tangent line with rise $m$ for run $1$ becomes a reflected line with rise $1$ for run $m$ , giving reciprocal slope $1/m$ .

The derivative must be evaluated at corresponding points. If $f(a)=b$ , then the inverse derivative at input $b$ depends on $f'(a)$ , not on $f'(b)$ . Keeping the point pair $(a,b)\leftrightarrow(b,a)$ visible prevents the usual substitution mistake.

Inverse functions exchange inputs and outputs, so their graphs reflect across the line $y=x$ . The slope at a reflected point becomes a reciprocal: a steep original graph produces a shallow inverse graph, provided the original slope is nonzero.

The rule is local. A function must actually be one-to-one on a suitable interval, and the derivative at the corresponding original point cannot vanish. Those conditions are what allow the inverse to behave like a smooth function nearby.

If $f$ is one-to-one and differentiable with $f'(f^{-1}(a))\ne0$ , then

\boxed{(f^{-1})'(a)=\frac1{f'(f^{-1}(a))}}.

Inverse graphs reflect across $y=x$ , which swaps rise and run. Their tangent slopes are reciprocal at corresponding points.

Guided walkthrough

Use inverse data without finding a formula

Suppose $f(2)=7$ and $f'(2)=5$ . Find $(f^{-1})'(7)$ .

Answer reveal

Worked solution

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Proof idea

The identity $f(f^{-1}(x))=x$ differentiates to

f'(f^{-1}(x))(f^{-1})'(x)=1.

Solving gives the inverse derivative formula.

Interactive checkinverse-function-01

If $f(4)=10$ and $f'(4)=8$ , find $(f^{-1})'(10)$ .

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Show hint

The inverse slope is the reciprocal of the original slope at the corresponding point.

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Application

Inverting a calibration curve

A laboratory instrument maps concentration $c$ to signal $S=f(c)$ . Suppose $f(4)=18$ and $f'(4)=2.5$ signal units per mg/L. The inverse calibration converts signal back to concentration, and

(f^{-1})'(18)=\frac{1}{2.5}=0.4\text{ mg/L per signal unit}.

A one-unit signal error near $18$ therefore produces about $0.4$ mg/L concentration error.

Optional advanced note

Local invertibility and nonzero slope

The condition $f'(a)\ne0$ does more than prevent division by zero in the inverse derivative formula. Under suitable continuity assumptions, it means the function is locally one-to-one near $a$ , so a differentiable inverse exists nearby. This one-dimensional fact becomes the Inverse Function Theorem for maps between higher-dimensional spaces.

After the explanation

Use the section idea

Reading lens

Track which variable depends on which and use reciprocal or logarithmic structure only where its conditions hold.

Mental model

Implicit equations constrain variables together; inverse functions exchange inputs and outputs; logarithms turn products and powers into sums.

Decision

Choose implicit, inverse, or logarithmic differentiation from the equation's representation, not from surface complexity.

Common trap

Dropping a y-prime factor, using a reciprocal slope at the wrong point, or ignoring domain restrictions.

Check yourself

Can you identify the correspondence point and all hidden dependencies before differentiating?

Interactive checkinverse-extra-01

If $f(2)=9$ and $f'(2)=6$ , find $(f^{-1})'(9)$ .

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Show hint

Use the reciprocal slope at the corresponding input.

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Source & rights

Original instruction with traceable references.

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