Derivatives of Hyperbolic Functions
Elegant Parallels with Sign Surprises
The derivatives of hyperbolic functions follow patterns remarkably similar to trigonometric derivatives, but with some unexpected sign differences. These differences aren’t random; they trace back to the fundamental identity $\cosh^2 x - \sinh^2 x = 1$ having a minus sign where the trig Pythagorean identity has a plus.
Mastering these derivatives opens doors to solving differential equations that model hanging cables, damped oscillations, and relativistic motion.
Before You Start: Quick Self-Check
1. What is $\frac{d}{dx}(e^x)$ and $\frac{d}{dx}(e^{-x})$?
Answer: $\frac{d}{dx}(e^x) = e^x$ and $\frac{d}{dx}(e^{-x}) = -e^{-x}$. These are essential for deriving hyperbolic derivatives.
2. State the chain rule for $\frac{d}{dx}[f(g(x))]$.
Answer: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$. If you need review, see Chain Rule.
3. What is $\cosh^2 x - \sinh^2 x$?
Answer: $\cosh^2 x - \sinh^2 x = 1$. This identity is used to simplify derivative results. Review Hyperbolic Identities if needed.
Prerequisite Map
Quick Reference
| Property | Value |
|---|---|
| Concept | Hyperbolic Functions |
| Chapter | 6.7 |
| Difficulty | Intermediate |
| Time | ~20 minutes |
Key Concepts
The Main Derivatives
$$\boxed{\frac{d}{dx}(\sinh x) = \cosh x \qquad \frac{d}{dx}(\cosh x) = \sinh x}$$
$$\frac{d}{dx}(\tanh x) = \text{sech}^2 x$$
Notice:
- $\sinh$ and $\cosh$ are derivatives of each other (no minus signs!)
- $\tanh$ differentiates to $\text{sech}^2$, paralleling $\frac{d}{dx}\tan x = \sec^2 x$
Complete Derivative Table
| Function | Derivative | Trig Analog |
|---|---|---|
| $\sinh x$ | $\cosh x$ | $\frac{d}{dx}\sin x = \cos x$ |
| $\cosh x$ | $\sinh x$ | $\frac{d}{dx}\cos x = -\sin x$ ← sign flip! |
| $\tanh x$ | $\text{sech}^2 x$ | $\frac{d}{dx}\tan x = \sec^2 x$ |
| $\text{csch } x$ | $-\text{csch } x \coth x$ | $\frac{d}{dx}\csc x = -\csc x \cot x$ |
| $\text{sech } x$ | $-\text{sech } x \tanh x$ | $\frac{d}{dx}\sec x = \sec x \tan x$ ← sign flip! |
| $\coth x$ | $-\text{csch}^2 x$ | $\frac{d}{dx}\cot x = -\csc^2 x$ |
Why the Derivatives Work
For $\sinh x$: $$\frac{d}{dx}(\sinh x) = \frac{d}{dx}\left(\frac{e^x - e^{-x}}{2}\right) = \frac{e^x + e^{-x}}{2} = \cosh x$$
For $\cosh x$: $$\frac{d}{dx}(\cosh x) = \frac{d}{dx}\left(\frac{e^x + e^{-x}}{2}\right) = \frac{e^x - e^{-x}}{2} = \sinh x$$
For $\tanh x$: Using the quotient rule: $$\frac{d}{dx}(\tanh x) = \frac{d}{dx}\left(\frac{\sinh x}{\cosh x}\right) = \frac{\cosh^2 x - \sinh^2 x}{\cosh^2 x} = \frac{1}{\cosh^2 x} = \text{sech}^2 x$$
The Chain Rule with Hyperbolic Functions
For composite functions, apply the chain rule as usual:
$$\frac{d}{dx}[\sinh(u)] = \cosh(u) \cdot \frac{du}{dx}$$ $$\frac{d}{dx}[\cosh(u)] = \sinh(u) \cdot \frac{du}{dx}$$ $$\frac{d}{dx}[\tanh(u)] = \text{sech}^2(u) \cdot \frac{du}{dx}$$
Memory Aid: Sign Patterns
Trig rule of thumb: When differentiating co-functions ($\cos$, $\cot$, $\csc$), you get a minus sign.
Hyperbolic modification: For $\cosh$ and $\text{sech}$, the expected trig pattern changes:
- $\frac{d}{dx}\cosh x = +\sinh x$ (not $-\sinh x$ like trig)
- $\frac{d}{dx}\text{sech } x = -\text{sech } x \tanh x$ (has a minus, unlike trig $\sec$)
Practice Problems
Find $\frac{d}{dx}(\sinh 5x)$.
Find $\frac{dy}{dx}$ if $y = \sinh(x^2 + 1)$.
Find $\frac{d}{dx}(x^2 \tanh x)$.
Find $\frac{d}{dx}[\ln(\cosh x)]$.
A hanging cable follows the catenary curve $y = c\cosh(x/c)$ where $c > 0$ is a constant.
(a) Show that this curve satisfies the differential equation $y'' = \frac{1}{c}\sqrt{1 + (y')^2}$.
(b) For a cable with $c = 8$, find the slope of the cable at the point where $x = 4$.
(c) At what value of $x$ does the slope of the cable equal 1?
CCI-Style Conceptual Questions
Question 1: Which of the following derivative formulas has a different sign pattern compared to its trigonometric counterpart?
(A) $\frac{d}{dx}\sinh x = \cosh x$ (B) $\frac{d}{dx}\tanh x = \text{sech}^2 x$ (C) $\frac{d}{dx}\cosh x = \sinh x$ (D) $\frac{d}{dx}\coth x = -\text{csch}^2 x$
Answer
(C) For trig: $\frac{d}{dx}\cos x = -\sin x$. For hyperbolic: $\frac{d}{dx}\cosh x = +\sinh x$. The hyperbolic version has no minus sign, unlike the trig case.
Question 2: If $f(x) = \tanh(x^2)$, then $f'(x)$ is:
(A) $\text{sech}^2(x^2)$ (B) $2x \cdot \text{sech}^2(x^2)$ (C) $2x \cdot \tanh(x^2)$ (D) $\text{sech}^2(2x)$
Answer
(B) By the chain rule: $\frac{d}{dx}\tanh(u) = \text{sech}^2(u) \cdot u'$, where $u = x^2$ and $u' = 2x$.
Applications
The Catenary Problem
A hanging cable satisfies the differential equation:
$$\frac{d^2y}{dx^2} = \frac{\rho g}{T}\sqrt{1 + \left(\frac{dy}{dx}\right)^2}$$
where $\rho$ is linear density, $g$ is gravity, and $T$ is tension at the lowest point.
The solution is $y = \frac{T}{\rho g}\cosh\left(\frac{\rho g x}{T}\right)$, a scaled hyperbolic cosine. The derivative formulas let us verify this solution:
$$y' = \sinh\left(\frac{\rho g x}{T}\right) \qquad y'' = \frac{\rho g}{T}\cosh\left(\frac{\rho g x}{T}\right)$$
Terminal Velocity
A falling object with air resistance has velocity modeled by:
$$v(t) = \sqrt{\frac{mg}{k}}\tanh\left(\sqrt{\frac{gk}{m}}t\right)$$
The derivative $\frac{dv}{dt}$ (acceleration) involves $\text{sech}^2$, which decreases as $t$ increases, modeling how acceleration diminishes as the object approaches terminal velocity.
Common Misconceptions
$\frac{d}{dx}[\cosh x] = -\sinh x$, by analogy with the circular trig identity $\frac{d}{dx}[\cos x] = -\sin x$.
This is the trig-as-algebra-symbols error. Hyperbolic and circular trigonometric functions have similar names but different derivative rules. For circular functions, $\frac{d}{dx}[\cos x] = -\sin x$; for hyperbolic functions, $\frac{d}{dx}[\cosh x] = \sinh x$ with no negative sign. This follows directly from the exponential definitions: $\frac{d}{dx}\!\left[\frac{e^x + e^{-x}}{2}\right] = \frac{e^x - e^{-x}}{2} = \sinh x$.
Mastery Checklist
Mental Model
The Symmetric Pair:
Think of $\sinh$ and $\cosh$ as a symmetric derivative pair: each is the derivative of the other. No minus signs appear between them, unlike $\sin$ and $\cos$ where $(\cos x)' = -\sin x$.
This makes $y = A\sinh x + B\cosh x$ particularly nice: its second derivative is just $y$ itself (when $m = 1$).
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|---|---|---|
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Last updated: 2026-01-22