MAT158H1/Unit 08
Unit 08 · Analysis 1A

Inverse Functions

When does a function have a continuous, differentiable inverse — and what is its derivative? The answer feeds directly into Unit 11, where $\exp$, the roots, and the inverse trig functions are all defined as inverses.

By the end you can
Why it matters & where it comes from
Why it exists

Many essential functions — $\exp$, $\sqrt[n]{x}$, $\arcsin$, $\arctan$ — are defined as inverses. We need to know they are differentiable and compute their derivatives without any explicit formula.

In the real world

The multivariable inverse and implicit function theorems (built on this one) underlie change-of-variables in integration, the Jacobians of robotics, and the solvability conditions in optimization.

Story & open questions

The 1-D version is elementary, but its generalizations are cornerstones of MAT257 and of differential geometry — the language of general relativity.

Existence and continuity of the inverse

Theorem

Let $f$ be continuous and strictly increasing on an interval $I$. Then its image $J=f(I)$ is an interval, $f:I\to J$ is a bijection, and $f^{-1}:J\to I$ is continuous and strictly increasing. (Same with "decreasing.")

Proof idea
Strict monotonicity gives injectivity outright, and the IVT (Unit 5) makes the image an interval: every value between two attained values is attained.
For continuity of $f^{-1}$ at $y_0=f(x_0)$: given $\varepsilon>0$, the image of $(x_0-\varepsilon,x_0+\varepsilon)$ is an interval containing $y_0$ strictly inside; any $\delta$ fitting inside it forces $|f^{-1}(y)-x_0|<\varepsilon$ whenever $|y-y_0|<\delta$. ∎
Monotone is the right hypothesis On an interval, a continuous injective function is automatically strictly monotone (an IVT argument) — so for continuous functions on intervals, "invertible" and "strictly monotone" coincide.

The derivative of the inverse

Theorem Inverse function theorem, 1-D

Let $f$ be continuous and strictly monotone on an interval, differentiable at $x_0$ with $f'(x_0)\ne0$, and $y_0=f(x_0)$. Then $f^{-1}$ is differentiable at $y_0$ and $$(f^{-1})'(y_0)=\frac{1}{f'(x_0)}.$$

Proof
Write $x=f^{-1}(y)$. As $y\to y_0$, continuity of $f^{-1}$ forces $x\to x_0$, and $y\ne y_0$ means $x\ne x_0$.
Then $\dfrac{f^{-1}(y)-f^{-1}(y_0)}{y-y_0}=\dfrac{x-x_0}{f(x)-f(x_0)} =\dfrac{1}{\dfrac{f(x)-f(x_0)}{x-x_0}}\longrightarrow\dfrac{1}{f'(x_0)}$, using $f'(x_0)\ne0$ and the reciprocal law. ∎
Example Why $f'(x_0)\ne0$ is needed

$f(x)=x^3$ is strictly increasing with $f'(0)=0$. Its inverse $y\mapsto y^{1/3}$ exists and is continuous, but has a vertical tangent at $0$ — not differentiable there. A horizontal tangent for $f$ always becomes a vertical one for $f^{-1}$.

Memory aid Reflect the graph across $y=x$: slopes invert, $m\mapsto1/m$. The theorem is the statement that this picture is honest — and the proof needed continuity of $f^{-1}$, which is why that theorem came first.

The payoff: roots and inverse trig

Example $n$th roots

$f(x)=x^n$ is strictly increasing on $(0,\infty)$, so $g(y)=y^{1/n}$ exists; the theorem gives, for $y>0$,

$g'(y)=\dfrac{1}{n\,x^{n-1}}$ with $x=y^{1/n}$, i.e. $g'(y)=\tfrac1n y^{\frac1n-1}$ — the power rule now holds for exponent $1/n$, and combining with the chain rule, for all rational exponents.
Example arcsin
$\sin$ is strictly increasing on $\left[-\tfrac\pi2,\tfrac\pi2\right]$; let $y=\arcsin x$, so $\sin y=x$.
The theorem gives $y'=\dfrac{1}{\cos y}$, and $\cos y=\sqrt{1-x^2}$ (positive on this branch), so $y'=\dfrac{1}{\sqrt{1-x^2}}$ for $|x|<1$. At $x=\pm1$, $\cos y=0$ and arcsine has vertical tangents — not differentiable there.
Common pitfalls

• The formula needs $f'(x_0)\ne0$; at a horizontal tangent the inverse has a vertical one.
• Evaluate $f'$ at $x_0=f^{-1}(y_0)$, not at $y_0$ — the most common slip in computations.
• Track the branch: $\arcsin$ lives on $\left[-\tfrac\pi2,\tfrac\pi2\right]$, which is what makes $\cos y\ge0$ and fixes the sign of the square root.

Practice

Original problems written for this guide in the style of Spivak's exercises. Write a full solution before opening the answer.

Practice 1 · arctan

Show $\dfrac{d}{dx}\arctan x=\dfrac{1}{1+x^2}$ for all real $x$.

Solution
$\tan y=x$ gives $y'=\dfrac{1}{\sec^2y}$ by the inverse rule, and $\sec^2y=1+\tan^2y=1+x^2$, which is never $0$ — so this works for every $x$.
Practice 2 · Numeric

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

Solution
$f$ is strictly increasing ($f'=3x^2+1>0$), $f(1)=2$, and $f'(1)=4$, so $(f^{-1})'(2)=\tfrac14$.
Practice 3 · Domains

Why is $f(x)=x^2$ on $\mathbb{R}$ not invertible, but on $[0,\infty)$ it is?

Answer
On $\mathbb{R}$ it is not injective ($f(\pm1)=1$). On $[0,\infty)$ it is strictly increasing, hence a bijection onto $[0,\infty)$ with inverse $\sqrt y$.
Practice 4 · Cube root

Use the inverse rule to differentiate $g(y)=y^{1/3}$ for $y\ne0$, and explain what happens at $y=0$.

Solution
With $f(x)=x^3$, $g'(y)=\dfrac{1}{3x^2}=\tfrac13y^{-2/3}$ where $x=y^{1/3}$. At $y=0$, $f'(0)=0$, and $g$ has a vertical tangent — not differentiable.
Practice 5 · Mixed function

Let $f(x)=2x+\cos x$. Show $f$ is invertible on $\mathbb{R}$ and compute $(f^{-1})'(1)$.

Solution
$f'(x)=2-\sin x\ge1>0$, so $f$ is strictly increasing, hence invertible. $f(0)=1$ and $f'(0)=2$, so $(f^{-1})'(1)=\tfrac12$.
Practice 6 · arccos

Derive $\dfrac{d}{dx}\arccos x=\dfrac{-1}{\sqrt{1-x^2}}$ for $|x|<1$.

Solution
$\cos y=x$ with $y\in(0,\pi)$ gives $-\sin y\,y'=1$. On that branch $\sin y=\sqrt{1-x^2}>0$, so $y'=-\dfrac{1}{\sqrt{1-x^2}}$.
Practice 7 · Theory

Prove: a continuous injective function on an interval is strictly monotone.

Hint
If not, there are $x_1<x_2<x_3$ with $f(x_2)$ not between $f(x_1)$ and $f(x_3)$. Then some value is attained on both $[x_1,x_2]$ and $[x_2,x_3]$ by the IVT, contradicting injectivity.
Practice 8 · Global bijection

Suppose $f$ is differentiable on $\mathbb{R}$ with $f'(x)\ge m>0$ for all $x$. Prove $f$ is a bijection from $\mathbb{R}$ onto $\mathbb{R}$.

Solution
$f$ is strictly increasing (MVT), hence injective. Also by the MVT, $f(x)\ge f(0)+mx\to\infty$ as $x\to\infty$ and $f(x)\le f(0)+mx\to-\infty$ as $x\to-\infty$; the IVT then hits every real value.

Expanded topic checklist

Course coverage. This unit is part of the complete MAT158H1 topic map. It should cover the following items before you move on:

Textbook/source alignment

Use the current UofT MAT158H1 calendar topics as the coverage spine, and pair lecture notes with a Spivak-style proof-calculus text for extra reading. The exact assigned textbook should still be checked against the instructor syllabus.

Core theorem map

The inverse-function spine is: prove one-to-one, identify the inverse domain and range, then differentiate through the reciprocal derivative rule.

Study routine. Always compute the derivative of the inverse at an output by first finding the corresponding input.

Deep practice extensions

Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.

Extension 1 · Inverse point

If $f(3)=10$ and $f\prime(3)=-2$, compute $(f^{-1})\prime(10)$ if the theorem applies.

Guide
It is $1/f\prime(3)=-1/2$.
Extension 2 · Monotone inverse

Explain why a continuous strictly increasing function on an interval has an inverse on its range.

Guide
Strict increase gives injectivity, so each output in the range has exactly one input; continuity controls interval behavior of the range.
Extension 3 · Derivative formula

Derive $(\ln x)\prime=1/x$ using $\ln$ as the inverse of $e^x$.

Guide
If $y=\ln x$, then $x=e^y$. The inverse derivative gives $dy/dx=1/e^y=1/x$.
Extension 4 · Domain check

Why is $x^2$ not invertible on all of $\mathbb{R}$ but invertible on $[0,\infty)$?

Guide
On all real numbers it is not one-to-one because $x$ and $-x$ have the same square. On $[0,\infty)$ it is strictly increasing.
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