v1 · padrão canônico

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)

Choose your door

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:

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

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

Ex. 56.1
What is the derivative of $y = \arcsin x$ ?
Solve online
Ex. 56.2Answer key
What is the derivative of $y = \arctan x$ ?
Solve online
Ex. 56.3Answer key
Differentiate $y = \arccos x$ by implicit differentiation. Explain why the result differs from $(\arcsin x)'$ only in sign.
Solve online
Ex. 56.4Answer key
Differentiate $y = \ln x$ by implicit differentiation.
Solve online
Ex. 56.5Answer key
Differentiate $y = \log_2 x$ .
Solve online
Ex. 56.6
What is the derivative of $y = a^x$ (with $a > 0$ , $a \neq 1$ )?
Solve online
Ex. 56.7Answer key
Differentiate $y = \text{arcsinh}\, x$ by implicit differentiation.
Solve online
Ex. 56.8
Differentiate $y = \text{arctanh}\, x$ (for $|x| < 1$ ).
Solve online
Ex. 56.9
Let $f(x) = x^3 + x$ . Given that $f(1) = 2$ , calculate $(f^{-1})'(2)$ .
Solve online
Ex. 56.10
Let $f(x) = e^x + x$ . Given that $f(0) = 1$ , calculate $(f^{-1})'(1)$ .
Solve online
Ex. 56.11
Calculate $\dfrac{d}{dx} 3^x$ and evaluate at $x = 1$ . Why does the power rule $nx^{n-1}$ not apply?
Solve online
Ex. 56.12
Calculate $\dfrac{d}{dx} 2^{x^2}$ .
Solve online
Ex. 56.13Answer key
Calculate $\dfrac{d}{dx}\arcsin(2x)$ .
Solve online
Ex. 56.14
Calculate $\dfrac{d}{dx}\arctan(x^2)$ .
Solve online
Ex. 56.15Answer key
Calculate $\dfrac{d}{dx}\arcsin(e^x)$ . What is the domain of this derivative?
Solve online
Ex. 56.16Answer key
Calculate $\dfrac{d}{dx}\arctan(\ln x)$ .
Solve online
Ex. 56.17
Calculate $\dfrac{d}{dx}\arcsin(x^3)$ .
Solve online
Ex. 56.18
Calculate $\dfrac{d}{dx}(\arctan x)^2$ .
Solve online
Ex. 56.19
Calculate $\dfrac{d}{dx}(\arcsin x + \arccos x)$ . Explain the result geometrically.
Solve online
Ex. 56.20
Calculate $\dfrac{d}{dx}\ln(\arctan x)$ and specify the domain.
Solve online
Ex. 56.21
Calculate $\dfrac{d}{dx}\!\left[x\arctan x - \dfrac{1}{2}\ln(1+x^2)\right]$ .
Solve online
Ex. 56.22
Calculate $\dfrac{d}{dx}\ln(\sec x + \tan x)$ .
Solve online
Ex. 56.23
Calculate $\dfrac{d}{dx}\!\left(\dfrac{\arctan x}{x}\right)$ .
Solve online
Ex. 56.24Answer key
Differentiate $y = \text{arcsec}\, x$ for $x > 1$ .
Solve online
Ex. 56.25Answer key
Calculate $\dfrac{d}{dx}\text{arccosh}(\ln x)$ . What is the domain?
Solve online
Ex. 56.26
Calculate $\dfrac{d}{dx}\!\left[(\arctan x)\ln(x^2+1)\right]$ .
Solve online
Ex. 56.27
Snell's Law. The angle of refraction satisfies $\theta_2 = \arcsin\!\left(\dfrac{n_1}{n_2}\sin\theta_1\right)$ . Calculate $d\theta_2/d\theta_1$ at $\theta_1 = 0$ .
Solve online
Ex. 56.28
GPS. The satellite's elevation angle is $\theta = \arctan(h/d)$ , where $h$ is altitude and $d$ is horizontal distance (fixed). Calculate the sensitivity $d\theta/dh$ .
Solve online
Ex. 56.29
Pendulum. The pendulum's angle satisfies $\theta = \arcsin(s/L)$ , where $s$ is the arc length and $L$ is the length. Calculate $d\theta/ds$ .
Solve online
Ex. 56.30
Use logarithmic differentiation to calculate $\dfrac{d}{dx} x^{\sin x}$ (for $x > 0$ ).
Solve online
Ex. 56.31
Use logarithmic differentiation to calculate $\dfrac{d}{dx} x^x$ (for $x > 0$ ).
Solve online
Ex. 56.32
Error Function. Let $F(x) = \displaystyle\int_0^x e^{-t^2}\,dt$ . Calculate $F'(x)$ by FTC and then determine $(F^{-1})'(0)$ .
Solve online
Ex. 56.33
Finance. The function $V(\sigma) = \text{BS}(\sigma)$ gives the price of an option as a function of volatility. The sensitivity of price to volatility is Vega. What is the sensitivity of implied volatility to market price, $d\sigma_{\text{imp}}/dV$ ?
Solve online
Ex. 56.34
Calculate $\dfrac{d}{dx}\arcsin(1/x)$ for $|x| > 1$ and compare with the derivative of $\text{arcsec}\, x$ .
Solve online
Ex. 56.35
Why must a function be strictly monotonic (and not just continuous) to have a well-defined inverse function?
Solve online
Ex. 56.36
What happens geometrically in the inverse derivative formula when $f'(a) = 0$ ?
Solve online
Ex. 56.37
Identity. Prove that $\arcsin x + \arccos x = \pi/2$ for all $x \in [-1, 1]$ using derivatives (show the difference is constant and evaluate at $x = 0$ ).
Solve online
Ex. 56.38
Lambert W Function. $W(x)$ satisfies $W(x)\,e^{W(x)} = x$ . Differentiate $W'(x)$ using implicit differentiation.
Solve online
Ex. 56.39
Use logarithmic differentiation to calculate $\dfrac{d}{dx}(\ln x)^{\ln x}$ for $x > 1$ .
Solve online
Ex. 56.40
Proof. Prove that $(f^{-1})'(y) = 1/f'(f^{-1}(y))$ using the identity $f(f^{-1}(y)) = y$ and the chain rule.
Solve online

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.

Rigorous Definition and Complete Table

Theorem of the Derivative of the Inverse Function

Proof via Chain Rule

Geometric Interpretation

Table of Derivatives of Inverse Functions

Chain Rule with Inverse Trig

Solved Examples

Sources