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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.
Answer: $\frac{d}{dx}(e^x) = e^x$ and $\frac{d}{dx}(e^{-x}) = -e^{-x}$. These are essential for deriving hyperbolic derivatives.
Answer: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$. If you need review, see Chain Rule.
Answer: $\cosh^2 x - \sinh^2 x = 1$. This identity is used to simplify derivative results. Review Hyperbolic Identities if needed.
| Property | Value |
|---|---|
| Concept | Hyperbolic Functions |
| Chapter | 6.7 |
| Difficulty | Intermediate |
| Time | ~20 minutes |
$$\boxed{\frac{d}{dx}(\sinh x) = \cosh x \qquad \frac{d}{dx}(\cosh x) = \sinh x}$$
$$\frac{d}{dx}(\tanh x) = \text{sech}^2 x$$
Notice:
| 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! |
| $\coth x$ | $-\text{csch}^2 x$ | $\frac{d}{dx}\cot x = -\csc^2 x$ |
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: $$\frac{d}{dx}(\tanh x) = \frac{d}{dx}\left(\frac{\sinh x}{\cosh x}\right) = \frac{\cosh^2 x - \sinh^2 x}{\cosh^2 x} = \frac{1}{\cosh^2 x} = \text{sech}^2 x$$
For composite functions, apply the chain rule as usual:
$$\frac{d}{dx}[\sinh(u)] = \cosh(u) \cdot \frac{du}{dx}$$ $$\frac{d}{dx}[\cosh(u)] = \sinh(u) \cdot \frac{du}{dx}$$ $$\frac{d}{dx}[\tanh(u)] = \text{sech}^2(u) \cdot \frac{du}{dx}$$
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:
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?
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) $\frac{d}{dx}\coth x = -\text{csch}^2 x$
(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) $\text{sech}^2(2x)$
(B) By the chain rule: $\frac{d}{dx}\tanh(u) = \text{sech}^2(u) \cdot u'$, where $u = x^2$ and $u' = 2x$.
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:
$$y' = \sinh\left(\frac{\rho g x}{T}\right) \qquad y'' = \frac{\rho g}{T}\cosh\left(\frac{\rho g x}{T}\right)$$
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.
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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Last updated: 2026-01-22