practicecalculus
Hyperbolic derivatives

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.

Hyperbolic derivatives Updated 2 min read

At a glance

\frac{d}{dx}\,\sinh x = \cosh x
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}}
On this page
  1. Proof from the definition
  2. Graph of sinh x and its derivative
  3. Hyperbolic vs circular derivatives
  4. Chain rule version and examples
  5. Derivative of the inverse, arsinh x
  6. Common mistakes

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 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} = \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 of y = sinh x and its derivative y = cosh x
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.

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: 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 and the derivative of tanh x. The hyperbolic functions page covers their 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 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}}.

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.