Write down (d)/(dx)( x). [1 mark]

State the identity linking ^2 x and ^2 x. [1 mark]

Using the log form, find arsinh 0. [1 mark]

CCEA CCEA A2 2021-style practice: Starting from the definitions x = (1)/(2)(e^x + e^-x) and x = (1)/(2)(e^x - e^-x), prove that ^2 x - ^2 x = 1.

Square each definition: ^2 x = (1)/(4)(e^x + e^-x)^2 = (1)/(4)(e^2x + 2 + e^-2x), ^2 x = (1)/(4)(e^x - e^-x)^2 = (1)/(4)(e^2x - 2 + e^-2x). Subtract: ^2 x - ^2 x = (1)/(4)[(e^2x + 2 + e^-2x) - (e^2x - 2 + e^-2x)] = (1)/(4)(4) = 1. Markers reward squaring both definitions correctly, the subtraction, and reaching 1.

CCEA CCEA A2 2019-style practice: Show that arsinh x = ln(x + sqrt(x^2 + 1)), and hence find (d)/(dx)(arsinh x).

Let y = arsinh x, so x = y = (1)/(2)(e^y - e^-y). Multiply by 2e^y: 2xe^y = e^2y - 1 e^2y - 2xe^y - 1 = 0. This is a quadratic in e^y; the positive root is e^y = x + sqrt(x^2 + 1), so y = ln(x + sqrt(x^2 + 1)), as required. Differentiate: ddx(arsinh x) = 1sqrt(x^2 + 1) (either from the log form or from (dy)/(dx) = (1)/( y) with y = sqrt(1 + x^2)). Markers reward the quadratic in e^y, choosing the positive root for the log form, and the derivative (1)/(sqrt(x^2 + 1)).

Northern IrelandFurther MathsSyllabus dot point

What are the hyperbolic functions, and how do you differentiate, integrate and invert them?

The definitions of the hyperbolic functions in terms of the exponential function, their identities, derivatives and integrals, and the logarithmic forms of the inverse hyperbolic functions.

A CCEA A2 Further Maths answer on the hyperbolic functions defined from the exponential function, the identities such as cosh squared minus sinh squared equals 1, their derivatives and integrals, and the logarithmic forms of the inverse hyperbolic functions.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-16

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

The hyperbolic functions are $\cosh x = \frac{1}{2}(e^x + e^{-x})$ , $\sinh x = \frac{1}{2}(e^x - e^{-x})$ and $\tanh x = \frac{\sinh x}{\cosh x}$ , with the key identity $\cosh^2 x - \sinh^2 x = 1$ . Their derivatives are $\frac{d}{dx}\sinh x = \cosh x$ and $\frac{d}{dx}\cosh x = \sinh x$ (no minus), and they integrate back to each other. The inverses have logarithmic forms such as $\operatorname{arsinh} x = \ln(x + \sqrt{x^2 + 1})$ , whose derivatives integrate surds like $\sqrt{x^2 \pm 1}$ via the substitutions $x = a\sinh u$ or $x = a\cosh u$ .

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What this dot point is asking

CCEA wants you to define the hyperbolic functions from the exponential function, use their identities, differentiate and integrate them, and derive and use the logarithmic forms of the inverse hyperbolic functions. These appear in integration and in solving equations.

The answer

Definitions

Identities

Derivatives and integrals

Note that $\frac{d}{dx}(\cosh x) = \sinh x$ has no minus sign, unlike the trigonometric $\cos$ .

Inverse hyperbolic functions

Integrating with a hyperbolic substitution

Find $\displaystyle\int \frac{1}{\sqrt{x^2 + 9}}\,dx$ .

step Choose the substitution

The $\sqrt{x^2 + a^2}$ form suggests $x = 3\sinh u$ , so $dx = 3\cosh u\,du$ and $x^2 + 9 = 9\sinh^2 u + 9 = 9\cosh^2 u$ .

step Substitute and simplify

\int \frac{3\cosh u}{\sqrt{9\cosh^2 u}}\,du = \int \frac{3\cosh u}{3\cosh u}\,du = \int 1\,du = u + c.

step Return to x

Since $x = 3\sinh u$ , $u = \operatorname{arsinh}\frac{x}{3}$ , so

\int \frac{1}{\sqrt{x^2 + 9}}\,dx = \operatorname{arsinh}\frac{x}{3} + c = \ln\left(x + \sqrt{x^2 + 9}\right) + c'.

Solving a hyperbolic equation

Solve $2\cosh x + \sinh x = 4$ , giving $x$ to three significant figures.

step Write in terms of exponentials

Using the definitions, $2\cosh x + \sinh x = (e^x + e^{-x}) + \tfrac{1}{2}(e^x - e^{-x}) = \tfrac{3}{2}e^x + \tfrac{1}{2}e^{-x}$ . Setting this equal to $4$ :

\tfrac{3}{2}e^x + \tfrac{1}{2}e^{-x} = 4.

step Form a quadratic in e to the x

Multiply through by $2e^x$ :

3e^{2x} + 1 = 8e^x \Rightarrow 3e^{2x} - 8e^x + 1 = 0.

Let

y = e^x

3y^2 - 8y + 1 = 0

, so

y = \dfrac{8 \pm \sqrt{64 - 12}}{6} = \dfrac{8 \pm \sqrt{52}}{6}

step Take logs of the positive roots

$y = 2.535$ or $y = 0.1315$ , both positive, so $x = \ln 2.535 = 0.930$ or $x = \ln 0.1315 = -2.03$ .

Examples in context

Example 1. The shape of a hanging chain. A flexible cable hanging under its own weight forms a catenary, $y = a\cosh\frac{x}{a}$ . The hyperbolic cosine is not a guess; it is the exact solution, which is why bridge and power-line engineers work with $\cosh$ .

Example 2. Special relativity. Velocities in relativity add through a quantity called rapidity, where $\tanh$ replaces the ordinary velocity ratio. The hyperbolic identities give the clean addition rule that ordinary fractions cannot.

Try this

Q1. Write down $\frac{d}{dx}(\sinh x)$ . [1 mark]

Cue. $\cosh x$ .

Q2. State the identity linking $\cosh^2 x$ and $\sinh^2 x$ . [1 mark]

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

Q3. Using the log form, find $\operatorname{arsinh} 0$ . [1 mark]

Cue. $\ln(0 + \sqrt{0 + 1}) = \ln 1 = 0$ .

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA A2 20215 marksStarting from the definitions

\cosh x = \frac{1}{2}(e^x + e^{-x})

and

\sinh x = \frac{1}{2}(e^x - e^{-x})

, prove that

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

Show worked answer →

Square each definition:

$\cosh^2 x = \frac{1}{4}(e^x + e^{-x})^2 = \frac{1}{4}(e^{2x} + 2 + e^{-2x}),$

$\sinh^2 x = \frac{1}{4}(e^x - e^{-x})^2 = \frac{1}{4}(e^{2x} - 2 + e^{-2x}).$

Subtract:

$\cosh^2 x - \sinh^2 x = \frac{1}{4}\big[(e^{2x} + 2 + e^{-2x}) - (e^{2x} - 2 + e^{-2x})\big] = \frac{1}{4}(4) = 1.$

Markers reward squaring both definitions correctly, the subtraction, and reaching $1$ .

CCEA A2 20196 marksShow that

\operatorname{arsinh} x = \ln\left(x + \sqrt{x^2 + 1}\right)

, and hence find

\frac{d}{dx}(\operatorname{arsinh} x)

Show worked answer →

Let $y = \operatorname{arsinh} x$ , so $x = \sinh y = \frac{1}{2}(e^y - e^{-y})$ . Multiply by $2e^y$ :

$2xe^y = e^{2y} - 1 \Rightarrow e^{2y} - 2xe^y - 1 = 0.$

This is a quadratic in $e^y$ ; the positive root is $e^y = x + \sqrt{x^2 + 1}$ , so $y = \ln\left(x + \sqrt{x^2 + 1}\right)$ , as required.

Differentiate: $\dfrac{d}{dx}(\operatorname{arsinh} x) = \dfrac{1}{\sqrt{x^2 + 1}}$ (either from the log form or from $\frac{dy}{dx} = \frac{1}{\cosh y}$ with $\cosh y = \sqrt{1 + x^2}$ ).

Markers reward the quadratic in $e^y$ , choosing the positive root for the log form, and the derivative $\frac{1}{\sqrt{x^2 + 1}}$ .

Related dot points

Sources & how we know this

CCEA GCE Further Mathematics specification — CCEA (2018)