practicecalculus
Hyperbolic derivatives

Derivative of cosh x: Formula, Proof and Examples

The derivative of cosh x is sinh x, with no minus sign. See the proof from exponentials, the hanging cable example, chain rule examples and a graph.

Hyperbolic derivatives Updated 2 min read

At a glance

\frac{d}{dx}\,\cosh x = \sinh x
Derivative
\sinh x
Definition
\cosh x = \dfrac{e^x + e^{-x}}{2}
Chain rule form
\dfrac{d}{dx}\cosh u = \sinh u\cdot u'
Key identity
\cosh^2 x - \sinh^2 x = 1
Inverse derivative
\dfrac{d}{dx}\operatorname{arcosh} x = \dfrac{1}{\sqrt{x^2 - 1}} for x > 1
On this page
  1. Proof from the definition
  2. Graph of cosh x and its derivative
  3. The hanging cable
  4. Chain rule version and examples
  5. Derivative of arcosh x
  6. Common mistakes

The derivative of \cosh x is \sinh x. Watch the sign: unlike the circular cosine, there is no minus.

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

Proof from the definition

The hyperbolic cosine is the average of e^x and e^{-x}:

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

Differentiate each piece. The derivative of eˣ is e^x, and the derivative of e^(−x) is -e^{-x}:

\frac{d}{dx}\,\frac{e^x + e^{-x}}{2} = \frac{e^x - e^{-x}}{2} = \sinh x

Graph of cosh x and its derivative

Graph of y = cosh x and its derivative y = sinh x
The hyperbolic cosine has a U shape with its minimum at (0, 1). Its slope, \sinh x, is negative on the left, zero at the bottom and positive on the right.

The derivative \sinh x is negative for x < 0 and positive for x > 0, so x = 0 is a minimum. The second derivative is \cosh x \ge 1 > 0, so the curve is concave up everywhere.

The hanging cable

A chain or cable hanging under its own weight takes the shape y = a\cosh\frac{x}{a}, called a catenary. Its slope at any point is

\frac{dy}{dx} = a\sinh\frac{x}{a}\cdot\frac{1}{a} = \sinh\frac{x}{a}

So the cable is flat at its lowest point and gets steeper toward the supports. The arc length formula also simplifies nicely, because 1 + \sinh^2 u = \cosh^2 u:

\sqrt{1 + \left(\frac{dy}{dx}\right)^2} = \sqrt{1 + \sinh^2\frac{x}{a}} = \cosh\frac{x}{a}

Chain rule version and examples

\frac{d}{dx}\cosh u = \sinh u\cdot u'
Function Derivative
\cosh(2x) 2\sinh(2x)
\cosh(x^2) 2x\sinh(x^2)
\cosh(\ln x) \frac{\sinh(\ln x)}{x} = \frac{1}{2}\left(1 - \frac{1}{x^2}\right)
\cosh^2 x 2\cosh x\sinh x = \sinh 2x
x\cosh x \cosh x + x\sinh x

In the third row, \sinh(\ln x) = \frac{x - 1/x}{2}, so dividing by x gives \frac{1}{2}\left(1 - \frac{1}{x^2}\right).

Worked example: checking the identity. Differentiate \cosh^2 x - \sinh^2 x:

2\cosh x\sinh x - 2\sinh x\cosh x = 0

A zero derivative means the expression is constant. At x = 0 it equals 1 - 0 = 1, which proves \cosh^2 x - \sinh^2 x = 1.

Derivative of arcosh x

For x > 1, the inverse hyperbolic cosine has derivative

\frac{d}{dx}\operatorname{arcosh} x = \frac{1}{\sqrt{x^2 - 1}}

Set \cosh y = x with y > 0, so \sinh y\,y' = 1 and \sinh y = \sqrt{\cosh^2 y - 1} = \sqrt{x^2 - 1}.

Common mistakes

  1. Writing -\sinh x. Copying the circular pattern is the most common error.
  2. Thinking \cosh x can be below 1. Its minimum is \cosh 0 = 1.
  3. Forgetting the inside derivative. \frac{d}{dx}\cosh(4x) = 4\sinh(4x).

See the derivative of sinh x, the derivative of tanh x and the hyperbolic functions page for graphs and identities.

Practice it now

Chain rule: e and ln

Solve real problems on this topic with instant feedback, hints and a step-by-step walkthrough for every answer. Free, no sign-up needed to start.

Practice this lesson

Frequently asked questions

What is the derivative of cosh x?

The derivative of \cosh x is \sinh x.

Is the derivative of cosh x equal to −sinh x?

No. That is the pattern for circular cosine, \frac{d}{dx}\cos x = -\sin x. For the hyperbolic cosine the answer is +\sinh x.

What is the derivative of cosh(2x)?

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

What is the second derivative of cosh x?

It is \cosh x again. Differentiating twice brings you back to the start.

Written by the PracticeCalculus team

We build interactive calculus lessons and write every page to be clear and correct. Spotted something off? Tell us and we will fix it.