# Derivative of sinh x: Formula, Proof and Examples

> The derivative of sinh x is cosh x. See the proof from the exponential definition, why there is no minus sign, chain rule examples and a graph.

Source: https://practicecalculus.com/derivatives/sinh-x
Updated: 2026-11-06

$$ \frac{d}{dx}\,\sinh x = \cosh x $$

## At a glance

- **Derivative:** $\cosh x$
- **Definition:** $\sinh x = \dfrac{e^x - e^{-x}}{2}$
- **Chain rule form:** $\dfrac{d}{dx}\sinh u = \cosh u\cdot u'$
- **Second derivative:** $\sinh x$
- **Inverse derivative:** $\dfrac{d}{dx}\operatorname{arsinh} x = \dfrac{1}{\sqrt{x^2 + 1}}$

## Key takeaways

- The derivative of $\sinh x$ is $\cosh x$.
- The proof follows straight from $\sinh x = \frac{e^x - e^{-x}}{2}$.
- Unlike $\sin x$ and $\cos x$, there are no minus signs: $\sinh$ and $\cosh$ are each other's derivatives.
- With the chain rule, $\frac{d}{dx}\sinh u = \cosh u\cdot u'$.

The **derivative of $\sinh x$ is $\cosh x$**. The hyperbolic sine behaves like the ordinary sine in many ways, but its derivative has no surprises with signs.

$$\frac{d}{dx}\,\sinh x = \cosh x$$

## Proof from the definition

The hyperbolic sine and cosine are defined with exponentials:

$$\sinh x = \frac{e^x - e^{-x}}{2} \qquad \cosh x = \frac{e^x + e^{-x}}{2}$$

Differentiate $\sinh x$ term by term. The [derivative of eˣ](/derivatives/e-to-the-x) is $e^x$, and the [derivative of e^(−x)](/derivatives/e-to-the-negative-x) is $-e^{-x}$:

$$\frac{d}{dx}\,\frac{e^x - e^{-x}}{2} = \frac{e^x - (-e^{-x})}{2} = \frac{e^x + e^{-x}}{2} = \cosh x$$

The two minus signs cancel, which is why the answer is $+\cosh x$.

## Graph of sinh x and its derivative

```graph
title: 'Graph of y = sinh x and its derivative y = cosh x'
caption: 'The slope of $\sinh x$ is never less than $1$, the value of $\cosh 0$. So $\sinh x$ is always increasing and passes through the origin at a 45° angle.'
x: [-3, 3]
y: [-6, 6]
functions:
  - expr: sinh(x)
    label: y = sinh x
  - expr: cosh(x)
    label: y = cosh x (derivative)
    dashed: true
points:
  - [0, 0, '(0, 0)']
  - [0, 1, 'slope 1']
```

Because $\cosh x \ge 1$ for every $x$, the graph of $\sinh x$ always rises, and its smallest slope is $1$ at the origin. Its tangent line at $0$ is $y = x$.

## Hyperbolic vs circular derivatives

| Circular | Derivative | Hyperbolic | Derivative |
| --- | --- | --- | --- |
| $\sin x$ | $\cos x$ | $\sinh x$ | $\cosh x$ |
| $\cos x$ | $-\sin x$ | $\cosh x$ | $\sinh x$ |
| $\tan x$ | $\sec^2 x$ | $\tanh x$ | $\operatorname{sech}^2 x$ |

The ordinary sine cycles with period four: $\sin \to \cos \to -\sin \to -\cos \to \sin$. The hyperbolic sine cycles with period two: $\sinh \to \cosh \to \sinh$. So $y = \sinh x$ solves $y'' = y$, while $y = \sin x$ solves $y'' = -y$.

## Chain rule version and examples

$$\frac{d}{dx}\sinh u = \cosh u\cdot u'$$

**Example 1.** $\frac{d}{dx}\sinh(3x) = 3\cosh(3x)$.

**Example 2.** $\frac{d}{dx}\sinh(x^2) = 2x\cosh(x^2)$.

**Example 3: product rule.** $\frac{d}{dx}\,x\sinh x = \sinh x + x\cosh x$.

**Example 4: a power.** $\frac{d}{dx}\sinh^2 x = 2\sinh x\cosh x = \sinh(2x)$, using the identity $\sinh 2x = 2\sinh x\cosh x$.

**Example 5: a log.** $\frac{d}{dx}\ln(\sinh x) = \frac{\cosh x}{\sinh x} = \coth x$ for $x > 0$.

## Derivative of the inverse, arsinh x

The inverse hyperbolic sine has a log formula, $\operatorname{arsinh} x = \ln\!\left(x + \sqrt{x^2 + 1}\right)$. Its derivative is

$$\frac{d}{dx}\operatorname{arsinh} x = \frac{1}{\sqrt{x^2 + 1}}$$

You can prove it like the [derivative of arcsin x](/derivatives/arcsin-x): set $\sinh y = x$, differentiate to get $\cosh y\,y' = 1$, and use $\cosh y = \sqrt{1 + \sinh^2 y} = \sqrt{1 + x^2}$. This result is why $\int\frac{dx}{\sqrt{x^2 + 1}} = \operatorname{arsinh} x + C$.

## Common mistakes

1. **Adding a minus sign.** $\frac{d}{dx}\sinh x = \cosh x$, not $-\cosh x$. Only circular cosine picks up a minus.
2. **Confusing $\sinh x$ with $\sin x$.** They share a pattern but are different functions with different graphs.
3. **Forgetting the chain rule.** $\frac{d}{dx}\sinh(5x) = 5\cosh(5x)$.

For the companion results, see the [derivative of cosh x](/derivatives/cosh-x) and the [derivative of tanh x](/derivatives/tanh-x). The [hyperbolic functions](/functions/hyperbolic-functions) page covers their graphs and identities.

## Frequently asked questions

### What is the derivative of sinh x?

The derivative of $\sinh x$ is $\cosh x$.

### What is the derivative of sinh(2x)?

By the chain rule it is $2\cosh(2x)$.

### Why is the derivative of sinh x not negative like cos?

Because $\sinh$ is built from exponentials, not from the unit circle. Differentiating $\frac{e^x - e^{-x}}{2}$ gives $\frac{e^x + e^{-x}}{2} = \cosh x$, with no sign change.

### What is the derivative of the inverse hyperbolic sine?

The derivative of $\operatorname{arsinh} x$ (also written $\sinh^{-1} x$) is $\frac{1}{\sqrt{x^2 + 1}}$.