Inverse Hyperbolic Functions
Why Inverse Hyperbolic Functions?
The integral $\int \frac{1}{\sqrt{1+x^2}}\,dx$ looks like it should involve $\arctan$ or $\arcsin$, but neither fits. The answer is $\sinh^{-1} x$, an inverse hyperbolic function.
Inverse hyperbolic functions complete the toolkit for integration. They appear whenever we need antiderivatives of expressions involving $\sqrt{x^2 \pm 1}$ or $\frac{1}{1-x^2}$. Even better, they have elegant logarithmic forms that make them easy to evaluate numerically.
Before You Start: Quick Self-Check
1. What is the domain and range of $\sinh x$?
Answer: Domain: $\mathbb{R}$. Range: $\mathbb{R}$. Since $\sinh$ is one-to-one, it has an inverse without restriction. If unsure, review Hyperbolic Function Definitions.
2. What is $\cosh^2 x - \sinh^2 x$?
Answer: $\cosh^2 x - \sinh^2 x = 1$. This identity is essential for deriving the derivative of $\sinh^{-1} x$. Review Hyperbolic Identities if needed.
3. Solve $e^y = 3$ for $y$.
Answer: $y = \ln 3$. Logarithm manipulation is essential for deriving the logarithmic forms of inverse hyperbolic functions.
Prerequisite Map
Quick Reference
| Property | Value |
|---|---|
| Concept | Hyperbolic Functions |
| Chapter | 6.7 |
| Difficulty | Advanced |
| Time | ~30 minutes |
Key Concepts
Definitions of Inverse Hyperbolic Functions
Since $\sinh$ is one-to-one on all of $\mathbb{R}$, it has an inverse:
$$y = \sinh^{-1} x \iff \sinh y = x$$
For $\cosh$, we restrict the domain to $[0, \infty)$ (where $\cosh$ is one-to-one):
$$y = \cosh^{-1} x \iff \cosh y = x \text{ and } y \geq 0$$
For $\tanh$, which maps $\mathbb{R}$ onto $(-1, 1)$:
$$y = \tanh^{-1} x \iff \tanh y = x$$
Domain and Range Summary
| Function | Domain | Range |
|---|---|---|
| $\sinh^{-1} x$ | $\mathbb{R}$ | $\mathbb{R}$ |
| $\cosh^{-1} x$ | $[1, \infty)$ | $[0, \infty)$ |
| $\tanh^{-1} x$ | $(-1, 1)$ | $\mathbb{R}$ |
The Logarithmic Forms
Since hyperbolic functions are defined using exponentials, their inverses can be expressed as logarithms:
$$\boxed{\sinh^{-1} x = \ln\left(x + \sqrt{x^2 + 1}\right)} \quad \text{for all } x$$
$$\boxed{\cosh^{-1} x = \ln\left(x + \sqrt{x^2 - 1}\right)} \quad \text{for } x \geq 1$$
$$\boxed{\tanh^{-1} x = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)} \quad \text{for } \vert x\vert < 1$$
Deriving the Logarithmic Form of $\sinh^{-1} x$
Let $y = \sinh^{-1} x$. Then $\sinh y = x$, which means:
$$\frac{e^y - e^{-y}}{2} = x$$
Multiply both sides by $2e^y$:
$$e^{2y} - 1 = 2xe^y$$
Rearrange as a quadratic in $e^y$:
$$(e^y)^2 - 2x(e^y) - 1 = 0$$
Apply the quadratic formula with $a=1$, $b=-2x$, $c=-1$:
$$e^y = \frac{2x \pm \sqrt{4x^2 + 4}}{2} = x \pm \sqrt{x^2 + 1}$$
Since $e^y > 0$ and $x - \sqrt{x^2 + 1} < 0$ for all $x$ (because $\sqrt{x^2+1} > \vert x\vert \geq x$), we must take the $+$ sign:
$$e^y = x + \sqrt{x^2 + 1}$$
Therefore:
$$y = \ln\left(x + \sqrt{x^2 + 1}\right)$$
Derivatives of Inverse Hyperbolic Functions
$$\boxed{\frac{d}{dx}(\sinh^{-1} x) = \frac{1}{\sqrt{1 + x^2}}}$$
$$\boxed{\frac{d}{dx}(\cosh^{-1} x) = \frac{1}{\sqrt{x^2 - 1}}} \quad \text{for } x > 1$$
$$\boxed{\frac{d}{dx}(\tanh^{-1} x) = \frac{1}{1 - x^2}} \quad \text{for } \vert x\vert < 1$$
Deriving $\frac{d}{dx}(\sinh^{-1} x)$ (Two Methods)
Method 1: Implicit Differentiation
Let $y = \sinh^{-1} x$. Then $\sinh y = x$.
Differentiate implicitly: $\cosh y \cdot \frac{dy}{dx} = 1$
So $\frac{dy}{dx} = \frac{1}{\cosh y}$.
From $\cosh^2 y - \sinh^2 y = 1$ and $\cosh y > 0$:
$$\cosh y = \sqrt{1 + \sinh^2 y} = \sqrt{1 + x^2}$$
Therefore: $\frac{d}{dx}(\sinh^{-1} x) = \frac{1}{\sqrt{1 + x^2}}$
Method 2: Differentiate the Logarithmic Form
$$\frac{d}{dx}\ln\left(x + \sqrt{x^2+1}\right) = \frac{1}{x + \sqrt{x^2+1}} \cdot \left(1 + \frac{x}{\sqrt{x^2+1}}\right)$$
$$= \frac{1}{x + \sqrt{x^2+1}} \cdot \frac{\sqrt{x^2+1} + x}{\sqrt{x^2+1}} = \frac{1}{\sqrt{x^2+1}}$$
Comparison with Inverse Trig Derivatives
| Inverse Hyperbolic | Derivative | Inverse Trig | Derivative |
|---|---|---|---|
| $\sinh^{-1} x$ | $\frac{1}{\sqrt{1+x^2}}$ | $\arctan x$ | $\frac{1}{1+x^2}$ |
| $\cosh^{-1} x$ | $\frac{1}{\sqrt{x^2-1}}$ | $\text{arcsec } x$ | $\frac{1}{\vert x\vert \sqrt{x^2-1}}$ |
| $\tanh^{-1} x$ | $\frac{1}{1-x^2}$ | N/A | N/A |
Notice that $\frac{d}{dx}\sinh^{-1} x$ has a square root while $\frac{d}{dx}\arctan x$ does not.
Practice Problems
Find the exact value of $\sinh^{-1} 0$.
Express $\cosh^{-1} 2$ as a logarithm and simplify.
Find $\frac{d}{dx}(\sinh^{-1} 3x)$.
Find $\frac{d}{dx}[\sinh^{-1}(e^x)]$.
Evaluate $\displaystyle\int_0^3 \frac{dx}{\sqrt{4+x^2}}$.
The logarithmic form of $\cosh^{-1} x$ is $\ln(x + \sqrt{x^2-1})$ for $x \geq 1$.
Verify that $\frac{d}{dx}[\ln(x + \sqrt{x^2-1})] = \frac{1}{\sqrt{x^2-1}}$ by direct differentiation.
The logarithmic form of $\tanh^{-1} x$ is $\tanh^{-1} x = \frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)$ for $\vert x\vert < 1$.
(a) Use this formula to find the exact value of $\tanh^{-1}\left(\frac{1}{3}\right)$.
(b) Verify your answer by showing that $\tanh\left(\text{your answer}\right) = \frac{1}{3}$.
CCI-Style Conceptual Questions
Question 1: The function $\sinh^{-1} x$ is defined for:
(A) $x \geq 0$ only (B) $x \geq 1$ only (C) $-1 < x < 1$ only (D) All real numbers $x$
Answer
(D) Since $\sinh: \mathbb{R} \to \mathbb{R}$ is a one-to-one function that achieves every real value, its inverse $\sinh^{-1}$ is defined for all real $x$.
Question 2: Why does $\cosh^{-1} x$ require $x \geq 1$?
(A) Because $\cosh x \geq 1$ for all $x$, so only values $\geq 1$ are in the range of $\cosh$ (B) Because the formula $\ln(x + \sqrt{x^2-1})$ requires $x \geq 1$ (C) Because $\cosh$ is not one-to-one for $x < 0$ (D) All of the above are correct explanations
Answer
(D) All three statements are true and related. The range of $\cosh$ is $[1, \infty)$, which is why only $x \geq 1$ can be inputs to $\cosh^{-1}$. The formula requires $x^2 - 1 \geq 0$, i.e., $x \geq 1$ (for the principal branch). And $\cosh$ is restricted to $x \geq 0$ to make it one-to-one.
Question 3 (Explain Why): The derivative $\frac{d}{dx}(\sinh^{-1} x) = \frac{1}{\sqrt{1+x^2}}$ is defined for all real $x$, but $\frac{d}{dx}(\cosh^{-1} x) = \frac{1}{\sqrt{x^2-1}}$ is only defined for $x > 1$. Why the difference?
Answer
The difference traces back to the domains of the inverse functions themselves:
$\sinh^{-1} x$ is defined for all $x \in \mathbb{R}$, so its derivative exists for all $x$. The expression $\sqrt{1+x^2}$ is always positive, so there are no restrictions.
$\cosh^{-1} x$ is only defined for $x \geq 1$. At $x = 1$, the derivative formula gives $\frac{1}{\sqrt{0}}$, which is undefined. The graph of $\cosh^{-1}$ has a vertical tangent at $x = 1$ (it starts from the point $(1, 0)$), so the derivative does not exist there. For $x > 1$, the derivative is well-defined.
Common Mistakes
| Mistake | Correct Approach |
|---|---|
| Confusing $\sinh^{-1} x$ with $\frac{1}{\sinh x}$ | $\sinh^{-1} x$ is the inverse function; $\frac{1}{\sinh x} = \text{csch } x$ |
| Using wrong formula sign: $\ln(x - \sqrt{x^2+1})$ | For $\sinh^{-1}$: use $+$ sign. The $-$ version gives a negative number. |
| Forgetting domain restrictions for $\cosh^{-1}$ and $\tanh^{-1}$ | $\cosh^{-1}$: $x \geq 1$. $\tanh^{-1}$: $\vert x\vert < 1$ |
| Missing chain rule factor | Always multiply by the derivative of the inner function |
Common Misconceptions
$\sinh^{-1}(x)$ has domain $[-1, 1]$, by analogy with $\sin^{-1}(x)$.
This is the concept-image-conflicts-definition error. The inverse circular sine $\sin^{-1}(x)$ requires $x \in [-1, 1]$ because $|\sin \theta| \leq 1$. However, $\sinh(t) = \frac{e^t - e^{-t}}{2}$ is unbounded in both directions, so $\sinh^{-1}(x)$ is defined for all real $x$. The logarithmic form $\sinh^{-1}(x) = \ln(x + \sqrt{x^2 + 1})$ confirms this, since $x + \sqrt{x^2+1} > 0$ for every real $x$.
Mastery Checklist
Mental Model
From Exponentials to Logarithms:
Hyperbolic functions are built from $e^x$ and $e^{-x}$. Their inverses must therefore be built from $\ln$.
When you set $\sinh y = x$, you’re solving an equation involving $e^y$ and $e^{-y}$. Multiplying through by $e^y$ converts this to a quadratic in $e^y$. The quadratic formula gives you $e^y$, and taking $\ln$ gives you $y$.
Integration Connection:
The derivatives of inverse hyperbolic functions are algebraic expressions:
- $\frac{1}{\sqrt{1+x^2}}$, $\frac{1}{\sqrt{x^2-1}}$, $\frac{1}{1-x^2}$
This means these algebraic expressions have inverse hyperbolic antiderivatives, a key fact for integration.
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|---|---|---|
| Hyperbolic Derivatives | Skills Index |
Last updated: 2026-01-23