Derivatives of Hyperbolic Functions: Formulas & Proofs

What Are The Derivatives Of Hyperbolic Functions?

The derivatives of hyperbolic functions are the rates of change of the six functions $\sinh x$, $\cosh x$, $\tanh x$, $\coth x$, $\operatorname{sech} x$ and $\operatorname{csch} x$. Hyperbolic sine and cosine differentiate into each other with no sign change, $\tanh x$ differentiates to $\operatorname{sech}^2 x$, and the three reciprocal functions each carry a minus sign.

The six functions are built from the exponential function. If you want the full background on their graphs and identities, the hyperbolic functions page covers it; here you only need the two definitions

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

where $e \approx 2.7183$ is Euler's number and $x$ is any real number. The other four are ratios of these two: $\tanh x = \frac{\sinh x}{\cosh x}$, $\coth x = \frac{\cosh x}{\sinh x}$, $\operatorname{sech} x = \frac{1}{\cosh x}$ and $\operatorname{csch} x = \frac{1}{\sinh x}$. One identity does most of the work below:

$$\cosh^2 x - \sinh^2 x = 1.$$

This article uses Leibniz notation, $\frac{d}{dx}$, throughout.

How Do You Derive The Derivatives Of Sinh And Cosh?

Differentiate the exponential definition term by term. The only rule you need is that the derivative of $e^{x}$ is $e^{x}$ and the derivative of $e^{-x}$ is $-e^{-x}$, both covered in derivatives of exponential functions.

For hyperbolic sine:

$$\frac{d}{dx}\sinh x = \frac{d}{dx}\left(\frac{e^{x} - e^{-x}}{2}\right)$$

$$= \frac{e^{x} - (-e^{-x})}{2}$$

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

$$= \cosh x.$$

For hyperbolic cosine:

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

$$= \frac{e^{x} + (-e^{-x})}{2}$$

$$= \frac{e^{x} - e^{-x}}{2}$$

$$= \sinh x.$$

Look at where the sign went. The minus that the chain rule produces from $e^{-x}$ flips the middle sign of each fraction, and that flip is what turns $\sinh$ into $\cosh$ and back again.

How Do You Get The Derivatives Of Tanh, Coth, Sech And Csch?

Each of the remaining four is a ratio, so the quotient rule (or the chain rule on a reciprocal) finishes the job. The identity $\cosh^2 x - \sinh^2 x = 1$ cleans up every numerator.

Hyperbolic Tangent

This is the derivation people ask for most often (on Quora, "How does one calculate the derivative of tanh(x)?"). Apply the quotient rule to $\tanh x = \frac{\sinh x}{\cosh x}$:

$$\frac{d}{dx}\tanh x = \frac{\cosh x \cdot \cosh x - \sinh x \cdot \sinh x}{\cosh^2 x}$$

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

$$= \frac{1}{\cosh^2 x}$$

$$= \operatorname{sech}^2 x.$$

Splitting the second line instead gives $1 - \frac{\sinh^2 x}{\cosh^2 x} = 1 - \tanh^2 x$, the same derivative written in terms of $\tanh x$ itself.

Hyperbolic Cotangent

The same rule on $\coth x = \frac{\cosh x}{\sinh x}$ gives

$$\frac{d}{dx}\coth x = \frac{\sinh^2 x - \cosh^2 x}{\sinh^2 x} = \frac{-1}{\sinh^2 x} = -\operatorname{csch}^2 x, \quad x \neq 0.$$

Hyperbolic Secant

Write $\operatorname{sech} x = (\cosh x)^{-1}$ and differentiate the outer power first:

$$\frac{d}{dx}\operatorname{sech} x = -(\cosh x)^{-2}\cdot \sinh x = -\frac{1}{\cosh x}\cdot\frac{\sinh x}{\cosh x} = -\operatorname{sech} x \tanh x.$$

Hyperbolic Cosecant

The same move on $(\sinh x)^{-1}$:

$$\frac{d}{dx}\operatorname{csch} x = -(\sinh x)^{-2}\cdot \cosh x = -\operatorname{csch} x \coth x, \quad x \neq 0.$$

The tangent picture is the geometric half of the table. At $x = 0$ the cosh curve sits at its lowest point, so its tangent is flat, and the formula agrees: $\sinh 0 = 0$. The tanh curve is steepest at the origin and flattens toward its asymptotes, which is exactly what $\operatorname{sech}^2 x$ does: it equals $1$ at $x = 0$ and shrinks toward $0$ as $|x|$ grows.

Why Do Hyperbolic Derivatives Have No Minus Sign Where Trig Does?

The short answer: $\cos x$ falls on $(0, \pi)$, so its derivative must be negative there, while $\cosh x$ rises for every $x > 0$, so its derivative must be positive. The algebra says the same thing, because the two minus signs in $\frac{d}{dx}e^{-x} = -e^{-x}$ cancel instead of surviving.

Put the two families side by side, with the trig rules from differentiation of trigonometric functions on the left:

Two rows differ in sign: cosine and secant. Everything else matches. That gives a memory rule that holds without exception:

• Trig: the three "co" functions ($\cos$, $\cot$, $\csc$) have negative derivatives.

• Hyperbolic: the three reciprocal functions ($\coth$, $\operatorname{sech}$, $\operatorname{csch}$) have negative derivatives, and the base three ($\sinh$, $\cosh$, $\tanh$) do not.

The sech row deserves a second look. We checked it twice while building this table, because $\sec x$ differentiates with a plus sign and it is very tempting to copy that across. The reason it flips: $\operatorname{sech} x = \frac{1}{\cosh x}$ shrinks as $x$ grows past $0$, so its slope there has to be negative.

How Does The Chain Rule Work With Hyperbolic Functions?

When the input is a function $u(x)$ instead of plain $x$, multiply the table entry by $\frac{du}{dx}$. That is the chain rule, and nothing about it changes for hyperbolic functions:

$$\frac{d}{dx}\sinh u = \cosh u \cdot \frac{du}{dx}, \qquad \frac{d}{dx}\cosh u = \sinh u \cdot \frac{du}{dx}, \qquad \frac{d}{dx}\tanh u = \operatorname{sech}^2 u \cdot \frac{du}{dx}.$$

For instance, with $u = x^{3}$:

$$\frac{d}{dx}\tanh\left(x^{3}\right) = 3x^{2}\operatorname{sech}^2\left(x^{3}\right).$$

The reciprocal three follow the same pattern, so $\frac{d}{dx}\operatorname{sech} u = -\operatorname{sech} u \tanh u \cdot \frac{du}{dx}$, and so on.

How Do You Find The Derivatives Of Inverse Hyperbolic Functions?

Set $y$ equal to the inverse function, rewrite it as a hyperbolic equation, and use implicit differentiation. The identity $\cosh^2 y - \sinh^2 y = 1$ then turns the answer back into $x$. This is the proof readers ask about when they search "What's the proof of the derivative of hyperbolic functions and inverse functions?"

Inverse Hyperbolic Sine

Let $y = \sinh^{-1} x$, so $\sinh y = x$. Differentiate both sides with respect to $x$:

$$\cosh y \cdot \frac{dy}{dx} = 1$$

$$\frac{dy}{dx} = \frac{1}{\cosh y}$$

$$= \frac{1}{\sqrt{1 + \sinh^2 y}}$$

$$= \frac{1}{\sqrt{1 + x^{2}}}.$$

The positive root is correct because $\cosh y \geq 1$ for every real $y$.

Inverse Hyperbolic Cosine

Let $y = \cosh^{-1} x$ with $x \geq 1$ and $y \geq 0$, so $\cosh y = x$. Then $\sinh y \cdot \frac{dy}{dx} = 1$, and because $y \geq 0$ makes $\sinh y \geq 0$,

$$\frac{dy}{dx} = \frac{1}{\sinh y} = \frac{1}{\sqrt{\cosh^2 y - 1}} = \frac{1}{\sqrt{x^{2} - 1}}, \quad x > 1.$$

Inverse Hyperbolic Tangent

Let $y = \tanh^{-1} x$ with $|x| < 1$. Then $\operatorname{sech}^2 y \cdot \frac{dy}{dx} = 1$, and $\operatorname{sech}^2 y = 1 - \tanh^2 y = 1 - x^{2}$, so

$$\frac{dy}{dx} = \frac{1}{1 - x^{2}}, \quad |x| < 1.$$

The Logarithm Route And The Full Table

There is a second route. Every inverse hyperbolic function has a closed form in terms of the natural logarithm, for example $\sinh^{-1} x = \ln\left(x + \sqrt{x^{2} + 1}\right)$ and $\tanh^{-1} x = \frac{1}{2}\ln\left(\frac{1 + x}{1 - x}\right)$. Differentiating those logarithms gives the same results, which is a useful check.

Two features of this table catch people out:

• Same formula, different domains. $\tanh^{-1} x$ and $\coth^{-1} x$ share the derivative $\frac{1}{1 - x^{2}}$, but one lives inside $(-1, 1)$ and the other outside it. The domain tells you which function you are looking at.

• A plus sign under the root. Compare $\frac{d}{dx}\sin^{-1} x = \frac{1}{\sqrt{1 - x^{2}}}$ from derivatives of inverse trigonometric functions with $\frac{d}{dx}\sinh^{-1} x = \frac{1}{\sqrt{1 + x^{2}}}$. The hyperbolic version has no domain restriction at all.

Why Do Hyperbolic Derivatives Matter? The Equation They Solve

"Functions that are their own second derivative."

That phrase is the reason these derivatives exist as a separate topic. Differentiate twice and you land back where you started:

$$\frac{d^{2}}{dx^{2}}\sinh x = \frac{d}{dx}\cosh x = \sinh x, \qquad \frac{d^{2}}{dx^{2}}\cosh x = \cosh x.$$

So both functions solve $\frac{d^{2}y}{dx^{2}} = y$, and the cycle repeats every two derivatives. Sine and cosine need four derivatives to return, because $\frac{d^{2}}{dx^{2}}\sin x = -\sin x$. The second derivative is exactly where the two families split.

Three consequences make this useful far beyond a formula sheet:

• A ready-made solution family. The general solution of $\frac{d^{2}w}{dz^{2}} - a^{2}w = 0$ is $w = A\cosh(az) + B\sinh(az)$, listed in the NIST Digital Library of Mathematical Functions, section 4.34. Starting conditions such as $w(0)$ and $w'(0)$ then read off directly, since $\cosh 0 = 1$ and $\sinh 0 = 0$.

• The hanging-chain problem. A cable hanging under its own weight satisfies a version of this equation, and its shape is $y = a\cosh\left(\frac{x}{a}\right)$. Leibniz, Huygens and Johann Bernoulli each found that curve in 1691, answering a challenge set by Jacob Bernoulli.

• Higher derivatives for free. The $n$th derivative of $\sinh x$ is $\sinh x$ when $n$ is even and $\cosh x$ when $n$ is odd, with no signs to track.

Examples Of Derivatives Of Hyperbolic Functions

Example 1

Differentiate $f(x) = 3\sinh x + 2\cosh x$ and find the slope at $x = 0$.

$$\frac{d}{dx}f(x) = 3\cosh x + 2\sinh x$$

At $x = 0$:

$$3\cosh 0 + 2\sinh 0 = 3(1) + 2(0) = 3$$

Final answer: $\frac{d}{dx}f(x) = 3\cosh x + 2\sinh x$, and the slope at $x = 0$ is $3$.

Example 2

Differentiate $y = \cosh\left(x^{2}\right)$.

The trig habit suggests copying the cosine rule, minus sign included, which produces $-2x\sinh\left(x^{2}\right)$.

Test it at $x = 1$. That expression gives $-2\sinh 1 \approx -2.3504$, a negative slope. But for $x > 0$ the inner value $x^{2}$ is increasing, and $\cosh$ increases for positive inputs, so $\cosh\left(x^{2}\right)$ must be rising at $x = 1$. A negative slope cannot be right.

The minus sign belongs to cosine, not to hyperbolic cosine. Use $\frac{d}{dx}\cosh u = \sinh u \cdot \frac{du}{dx}$ with $u = x^{2}$:

$$\frac{dy}{dx} = \sinh\left(x^{2}\right)\cdot 2x$$

$$= 2x\sinh\left(x^{2}\right)$$

Check at $x = 1$: $2\sinh 1 \approx 2.3504$, positive, as it should be.

Final answer: $\frac{dy}{dx} = 2x\sinh\left(x^{2}\right)$.

Example 3

Differentiate $y = x^{2}\sinh x$.

This is a product, so use the product rule:

$$\frac{dy}{dx} = \frac{d}{dx}\left(x^{2}\right)\cdot\sinh x + x^{2}\cdot\frac{d}{dx}\sinh x$$

$$= 2x\sinh x + x^{2}\cosh x$$

Final answer: $\frac{dy}{dx} = 2x\sinh x + x^{2}\cosh x$.

Example 4

Differentiate $y = \ln(\cosh x)$.

This is a Quora favourite ("What's the derivative of log(cosh(x))?"). The first instinct is to write $\frac{1}{\cosh x}$ and stop, which drops the inner derivative. The outer function is $\ln u$ with $u = \cosh x$:

$$\frac{dy}{dx} = \frac{1}{\cosh x}\cdot\frac{d}{dx}\cosh x$$

$$= \frac{\sinh x}{\cosh x}$$

$$= \tanh x$$

Final answer: $\frac{dy}{dx} = \tanh x$. Read backwards, this also says that an antiderivative of $\tanh x$ is $\ln(\cosh x)$.

Example 5

Differentiate $y = \tanh^{-1}(\sin x)$ wherever it is defined.

The function needs $|\sin x| < 1$, so $\cos x \neq 0$. With $u = \sin x$:

$$\frac{dy}{dx} = \frac{1}{1 - \sin^{2} x}\cdot\cos x$$

$$= \frac{\cos x}{\cos^{2} x}$$

$$= \frac{1}{\cos x}$$

$$= \sec x$$

Final answer: $\frac{dy}{dx} = \sec x$, for every $x$ with $\cos x \neq 0$.

Example 6

A cable hangs in the shape $y = 20\cosh\left(\frac{x}{20}\right)$, with $x$ and $y$ in metres. Find its slope and its angle above the horizontal at $x = 10$.

Differentiate with $u = \frac{x}{20}$:

$$\frac{dy}{dx} = 20\sinh\left(\frac{x}{20}\right)\cdot\frac{1}{20}$$

$$= \sinh\left(\frac{x}{20}\right)$$

At $x = 10$:

$$\frac{dy}{dx} = \sinh(0.5) \approx 0.5211$$

The angle $\theta$ satisfies $\tan\theta = 0.5211$, so

$$\theta \approx 27.5238^{\circ}$$

Final answer: the slope is $\sinh(0.5) \approx 0.5211$, an angle of about $27.5238^{\circ}$ above the horizontal.

Where Are Derivatives Of Hyperbolic Functions Used In Real Life?

Derivatives of hyperbolic functions show up wherever a quantity grows or levels off in a balanced, two-sided way. A few places they do real work:

• Cables, chains and arches. The slope of $y = a\cosh\left(\frac{x}{a}\right)$ is $\sinh\left(\frac{x}{a}\right)$, and the identity $1 + \sinh^2 u = \cosh^2 u$ is what makes the cable's length computable. OpenStax Calculus Volume 1, section 6.9 works through a cable-length problem this way.

• Falling with air resistance. A body falling with drag proportional to $v^{2}$ obeys $\frac{dv}{dt} = g - kv^{2}$, where $g$ is gravitational acceleration and $k$ is the drag constant per unit mass. The solution starting from rest is $v(t) = \sqrt{\frac{g}{k}}\tanh\left(\sqrt{gk},t\right)$. Differentiating gives $\frac{dv}{dt} = g\operatorname{sech}^2\left(\sqrt{gk},t\right)$, which shrinks to $0$, so the speed levels off at $\sqrt{\frac{g}{k}}$, the terminal velocity.

• Neural networks. The $\tanh$ function is a common activation in machine learning. Its derivative $1 - \tanh^2 x$ can be computed from the output alone, which saves work when a network adjusts its weights.

Where Do Students Trip Up On Derivatives Of Hyperbolic Functions?

The mistakes here are rarely about calculus. They come from importing trig habits, or from reading notation the wrong way.

Mistake 1: Dropping Every Minus Sign After Learning That Cosh Has None

Where it slips in: Differentiating $\operatorname{sech} x$, $\operatorname{csch} x$ or $\coth x$ right after learning that $\frac{d}{dx}\cosh x = \sinh x$.

Don't do this: Write $\frac{d}{dx}\operatorname{sech} x = \operatorname{sech} x \tanh x$ because "hyperbolic derivatives have no minus signs".

The correct way: Only the base three ($\sinh$, $\cosh$, $\tanh$) are sign-free. The reciprocals all carry a minus: $\frac{d}{dx}\operatorname{sech} x = -\operatorname{sech} x \tanh x$.

The student who memorised "no minus signs in hyperbolic land" as a single slogan gets cosh right every time, then over-applies it to sech. The fix is the reciprocal rule from the comparison section, not more drilling.

Mistake 2: Reading Sinh Inverse As One Over Sinh

Where it slips in: Any problem written with $\sinh^{-1} x$ instead of $\operatorname{arsinh} x$.

Don't do this: Treat $\sinh^{-1} x$ as $\frac{1}{\sinh x} = \operatorname{csch} x$ and differentiate to $-\operatorname{csch} x\coth x$.

The correct way: The superscript $-1$ means the inverse function, so $\frac{d}{dx}\sinh^{-1} x = \frac{1}{\sqrt{x^{2} + 1}}$. The reciprocal has its own name, $\operatorname{csch} x$.

Mistake 3: Ignoring The Domain Of An Inverse Hyperbolic Function

Where it slips in: Composite inputs such as $\cosh^{-1}(\cos x)$, a question that comes up on Quora as "How to find the derivative of cosh-1 (cos x)".

Don't do this: Apply the formula blindly to get $\frac{-\sin x}{\sqrt{\cos^{2} x - 1}}$ and report it as the derivative.

The correct way:Check the domain first. $\cosh^{-1} u$ needs $u \geq 1$, but $\cos x \leq 1$, so $\cosh^{-1}(\cos x)$ is defined only at isolated points where $\cos x = 1$. The square root above is of a negative number almost everywhere, so there is no real derivative to find.

The rusher writes the formula before asking whether the function exists, and this problem is designed to catch exactly that habit. The same check separates $\tanh^{-1} x$ (inside $(-1, 1)$) from $\coth^{-1} x$ (outside it).

Swapping a familiar function for a hyperbolic one has a famous real-world history. Galileo claimed that a hanging chain forms a parabola, and work by Joachim Jungius, published in 1669, showed that it does not. The true curve, the cosh-shaped catenary, had to wait until 1691, as the MacTutor history of the catenary records. A parabola and a catenary look close near the bottom, which is exactly why the substitution is so easy to make and so hard to spot.

Practice Problems On Derivatives Of Hyperbolic Functions

Find $\frac{dy}{dx}$ for each function.

1. $y = \sinh(5x)$

2. $y = x\cosh x$

3. $y = \operatorname{sech}(2x)$

4. $y = \tanh^{2} x$

5. $y = \sinh^{-1}\left(\frac{x}{2}\right)$

6. $y = x\tanh x$

Answer to Question 1: $5\cosh(5x)$

Answer to Question 2: $\cosh x + x\sinh x$

Answer to Question 3: $-2\operatorname{sech}(2x)\tanh(2x)$

Answer to Question 4: $2\tanh x\operatorname{sech}^{2} x$

Answer to Question 5: $\frac{1}{\sqrt{x^{2} + 4}}$

Answer to Question 6: $\tanh x + x\operatorname{sech}^{2} x$

Key Takeaways

• The derivatives of hyperbolic functions start from two facts: $\sinh x$ differentiates to $\cosh x$, and $\cosh x$ differentiates to $\sinh x$, with no minus sign.

• The derivative of $\tanh x$ is $\operatorname{sech}^2 x$, which also equals $1 - \tanh^2 x$.

• Only the three reciprocal functions, $\coth x$, $\operatorname{sech} x$ and $\operatorname{csch} x$, carry a minus sign in their derivatives.

• Inverse hyperbolic derivatives come from implicit differentiation, and each one is valid only on its own domain.

• Both $\sinh x$ and $\cosh x$ equal their own second derivative, which is why they solve hanging-cable and drag problems.

Your Next Move

Work through practice problems 1 to 6 without looking at the answers; if problem 5 trips you up, return to the section on inverse hyperbolic functions and redo the $\sinh^{-1} x$ derivation with $u = \frac{x}{2}$. Book a free Bhanzu demo class:

Read More

• Table of derivatives for every standard function

• Inverse functions and differentiation

• Logarithmic differentiation

• Derivatives of logarithmic functions

• Arc length of a curve