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Lesson 56 — Derivatives of Inverse Functions

The Inverse Function Theorem and differentiation of arcsin, arccos, arctan, ln, log_a, a^x, and inverse hyperbolics via implicit differentiation.

Used in: Advanced HS Math Year 2 · Japanese Math III equiv. chap. 3 · German Analysis Grundkurs/Leistungskurs equiv. · IB Math HL topic 6

(f^{-1})'(b) = \dfrac{1}{f'(a)}, \quad b = f(a)

The derivative of the inverse function: the slope of $f^{-1}$ at point $b$ is the reciprocal of the slope of $f$ at point $a$ , where $b = f(a)$ . This formula generates the derivatives of $\arcsin$ , $\arctan$ , $\ln$ , and all elementary inverse functions.

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Rigorous notation, full derivation, hypotheses

Rigorous Definition and Complete Table

Theorem of the Derivative of the Inverse Function

"If $f$ is a differentiable, one-to-one function with $f(a) = b$ and $f'(a) \neq 0$ , then $f^{-1}$ is differentiable at $b$ and $(f^{-1})'(b) = 1/f'(a)$ ." — Active Calculus §2.6, Theorem 2.6.2

Proof via Chain Rule

From the identity $f(f^{-1}(y)) = y$ , differentiating both sides with respect to $y$ using the chain rule:

$f'(f^{-1}(y)) \cdot (f^{-1})'(y) = 1$

Since $f'(f^{-1}(y)) \neq 0$ by hypothesis, we can divide:

$\boxed{(f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))}}$

Geometric Interpretation

The graph of $f^{-1}$ is the reflection of the graph of $f$ across the line $y = x$ . A tangent line with slope $m$ on the graph of $f$ at point $(a, b)$ becomes a tangent line with slope $1/m$ on the graph of $f^{-1}$ at point $(b, a)$ — the reflection swaps the roles of $\Delta x$ and $\Delta y$ .

Reflection across the diagonal y=x transforms slope m into 1/m. Point (a, b) on f becomes (b, a) on f⁻¹.

Table of Derivatives of Inverse Functions

Function	Domain	Derivative
$\arcsin x$	$(-1, 1)$	$\dfrac{1}{\sqrt{1 - x^2}}$
$\arccos x$	$(-1, 1)$	$-\dfrac{1}{\sqrt{1 - x^2}}$
$\arctan x$	$\mathbb{R}$	$\dfrac{1}{1 + x^2}$
$\text{arccot}\, x$	$\mathbb{R}$	$-\dfrac{1}{1 + x^2}$
$\text{arcsec}\, x$	$\vert x \vert > 1$	$\dfrac{1}{\vert x \vert\sqrt{x^2 - 1}}$
$\text{arccsc}\, x$	$\vert x \vert > 1$	$-\dfrac{1}{\vert x \vert\sqrt{x^2 - 1}}$
$\ln x$	$x > 0$	$\dfrac{1}{x}$
$\log_a x\;(a>0,\,a\neq1)$	$x > 0$	$\dfrac{1}{x \ln a}$
$\text{arcsinh}\, x$	$\mathbb{R}$	$\dfrac{1}{\sqrt{x^2 + 1}}$
$\text{arccosh}\, x$	$x > 1$	$\dfrac{1}{\sqrt{x^2 - 1}}$
$\text{arctanh}\, x$	$	x

"In general, there is a formula for the derivative of $a^x$ for any $a > 0$ with $a \neq 1$ : $\frac{d}{dx}(a^x) = a^x \ln a$ . This formula is a special case of the chain rule applied to $a^x = e^{x \ln a}$ ." — OpenStax Calculus Volume 1 §3.7

Chain Rule with Inverse Trig

For a differentiable $u = g(x)$ :

$\frac{d}{dx}\arcsin(g(x)) = \frac{g'(x)}{\sqrt{1 - g(x)^2}}, \qquad \frac{d}{dx}\arctan(g(x)) = \frac{g'(x)}{1 + g(x)^2}$

Solved Examples

Example— 1· Derivative of arctan via implicit differentiation (fundamental)

Problem: Differentiate $(\arctan x)'$ starting from the definition via implicit differentiation.

Strategy: Use the standard method: write the inverse relation $\tan y = x$ , differentiate implicitly, and rewrite the result in terms of $x$ .

Solution:

Let $y = \arctan x$ . By definition of the inverse, $\tan y = x$ with $y \in (-\pi/2,\, \pi/2)$ .
Differentiate both sides with respect to $x$ : $\sec^2 y \cdot \frac{dy}{dx} = 1$
Isolate $dy/dx$ : $\frac{dy}{dx} = \frac{1}{\sec^2 y}$
Rewrite $\sec^2 y$ : using the identity $\sec^2 y = 1 + \tan^2 y$ and substituting $\tan y = x$ : $\sec^2 y = 1 + x^2$
Therefore: $\frac{dy}{dx} = \frac{1}{1 + x^2}$

Verification: At $x = 0$ : $(f^{-1})'(0) = 1/(1+0) = 1$ . Since $f(x) = \tan x$ has $f'(0) = \sec^2 0 = 1$ , and $1/f'(0) = 1$ . It checks out.

Source. Active Calculus §2.6 — Boelkins · CC-BY-NC-SA.

Example— 2· Derivative of ln via inverse of exp (fundamental)

Problem: Differentiate $(\ln x)'$ starting from $e^y = x$ .

Strategy: $\ln$ is the inverse of $e^x$ . Apply implicit differentiation and use $e^y = x$ .

Solution:

Let $y = \ln x$ . Then $e^y = x$ (definition of the natural logarithm).
Differentiate both sides with respect to $x$ : $e^y \cdot \frac{dy}{dx} = 1$
Isolate $dy/dx$ : $\frac{dy}{dx} = \frac{1}{e^y}$
Substitute $e^y = x$ : $\frac{dy}{dx} = \frac{1}{x}$

Result: $(\ln x)' = 1/x$ for $x > 0$ .

Derivative of $\log_a x$ : Since $\log_a x = \ln x / \ln a$ : $(\log_a x)' = \frac{1}{x \ln a}$

Verification: At $x = 1$ : $(\ln x)'\big|_{x=1} = 1$ . The tangent line to $y = \ln x$ at $(1, 0)$ has slope 1. Checks out.

Source. OpenStax Calculus Volume 1 §3.7 · CC-BY-NC-SA.

Example— 3· Chain rule with arcsin — composition arcsin(2x)

Problem: Calculate $\dfrac{d}{dx}\arcsin(2x)$ .

Strategy: Apply the formula $(\arcsin u)' = u'/\sqrt{1-u^2}$ with $u = 2x$ .

Solution:

Identify $u = 2x$ , so $u' = 2$ .
Apply the chain rule: $\frac{d}{dx}\arcsin(2x) = \frac{2}{\sqrt{1 - (2x)^2}} = \frac{2}{\sqrt{1 - 4x^2}}$
The result is valid for $|2x| < 1$ , or $x \in (-1/2,\, 1/2)$ .

Verification: For $x = 0$ : derivative $= 2/\sqrt{1} = 2$ . By finite difference: $\arcsin(2 \cdot 0.01) \approx 0.020$ , divided by $0.01$ gives $\approx 2.00$ . Checks out.

Source. Active Calculus §2.6, Exercise 2.6.2 — Boelkins · CC-BY-NC-SA.

Example— 4· Inverse theorem without explicit formula — calculate (f^{-1})'(b)

Problem: Let $f(x) = x^3 + 2x + 1$ . Calculate $(f^{-1})'(4)$ .

Strategy: We don't know $f^{-1}$ explicitly, but we can use $(f^{-1})'(b) = 1/f'(f^{-1}(b))$ . We need to find $a$ such that $f(a) = 4$ .

Solution:

Find $a$ such that $f(a) = 4$ : $a^3 + 2a + 1 = 4 \implies a^3 + 2a - 3 = 0$

Test $a = 1$ : $1 + 2 - 3 = 0$ . So $a = 1$ , meaning $f(1) = 4$ .

Calculate $f'(x) = 3x^2 + 2$ .
Evaluate at $a = 1$ : $f'(1) = 3(1)^2 + 2 = 5$ .
Apply the theorem: $(f^{-1})'(4) = \frac{1}{f'(1)} = \frac{1}{5}$

Verification: $f$ is strictly increasing ( $f'(x) = 3x^2 + 2 > 0$ for all $x$ ), so $f^{-1}$ exists. At $b = 4$ , the slope of $f^{-1}$ is $1/5$ .

Source. Active Calculus §2.6, Exercise 2.6.4 — Boelkins · CC-BY-NC-SA.

Example— 5· Logarithmic differentiation — derivative of x^x

Problem: Differentiate $y = x^x$ for $x > 0$ .

Strategy: $x^x$ is neither a fixed power nor a fixed exponential — use logarithmic differentiation: take $\ln$ of both sides and differentiate.

Solution:

Take $\ln$ of both sides: $\ln y = x \ln x$
Differentiate implicitly with respect to $x$ : $\frac{1}{y} \cdot \frac{dy}{dx} = \ln x + x \cdot \frac{1}{x} = \ln x + 1$
Isolate $dy/dx$ by multiplying by $y = x^x$ : $\frac{dy}{dx} = x^x (\ln x + 1)$

Particular Case: At $x = 1$ : $dy/dx = 1^1(\ln 1 + 1) = 1 \cdot 1 = 1$ . The tangent line to $y = x^x$ at $(1, 1)$ has slope 1.

Generalization: For $y = f(x)^{g(x)}$ , logarithmic differentiation gives: $\frac{dy}{dx} = f(x)^{g(x)} \left(g'(x)\ln f(x) + \frac{g(x) f'(x)}{f(x)}\right)$

Source. OpenStax Calculus Volume 1 §3.7 · CC-BY-NC-SA.

Exercise list

40 exercises · 10 with worked solution (25%)

26 8 3 2 1

Sources

Active Calculus — Boelkins · 2024 · §2.6 "Derivatives of Inverse Functions" · CC-BY-NC-SA. Primary source. Free online section with discovery activities.
Calculus Volume 1 — OpenStax · 2016 · §3.7 "Derivatives of Inverse Functions" · CC-BY-NC-SA. Complete table, examples of logarithmic differentiation.
APEX Calculus — Hartman et al. · 2024 · v5 · §2.7 & §6.6 · CC-BY-NC. Free PDF. Inverse hyperbolics and advanced compositions.