Hyperbolic Functions A-Level Further Maths Guide

A sixth-form student can often recognise every symbol in a hyperbolic functions question and still have no idea where to begin. The notation resembles trigonometry, but the graphs, identities and inverse functions behave differently.

This guide to hyperbolic functions A-Level Further Maths gives you a practical route through the topic: start with exponential definitions, learn one central identity, connect differentiation and integration, then practise choosing the right method in exam questions.

Imagine a student called Alex meeting the topic for the first time. At first, Alex treats $\sinh x$ as though it were simply another version of $\sin x$. Progress begins when Alex notices something more useful: hyperbolic functions are built from exponentials. That observation makes almost the entire topic more manageable.

What you need to know

A secure revision checklist includes:

• the definitions of $\sinh x$, $\cosh x$ and $\tanh x$
• the shapes, domains and ranges of their graphs
• the identity ${\cosh }^{2}x-{\sinh }^{2}x=1$
• differentiation and integration involving hyperbolic functions
• inverse hyperbolic functions and their logarithmic forms
• solving equations and proving identities
• recognising when a hyperbolic substitution may help

These skills form part of core pure content in specifications such as Edexcel A Level Further Mathematics. You can find the relevant lessons and board resources through the Edexcel A Level Further Maths revision hub.

What are hyperbolic functions?

The three main hyperbolic functions are defined using $e^x$ and $e^{-x}$:

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

and

$$\tanh x=\frac{\sinh x}{\cosh x}=\frac{e^x-e^{-x}}{e^x+e^{-x}}.$$

These definitions are your safety net. If you forget an identity or need to prove a result, returning to the exponential forms will usually give you a way forward.

The graphs also reveal important facts:

• $\sinh x$ is odd, so $\sinh (-x)=-\sinh x$. Its domain and range are both $\mathbb{R}$.
• $\cosh x$ is even, so $\cosh (-x)=\cosh x$. Its minimum value is $1$, giving the range $[1,\infty )$.
• $\tanh x$ is odd and has horizontal asymptotes $y=1$ and $y=-1$. Its range is $(-1,1)$.

Alex's first useful exam habit is therefore to check whether a proposed answer respects the range. An equation such as $\cosh x=-2$ has no real solution, however convincing the algebra may look.

The central hyperbolic identity

The identity you will use most often is

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

Notice the minus sign. This is not the trigonometric identity ${\cos }^{2}x+{\sin }^{2}x=1$ with longer function names.

It follows directly from the definitions:

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

Expanding gives

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

Dividing the identity by ${\cosh }^{2}x$ gives another useful result:

$$1-{\tanh }^{2}x={\mathrm{sech}}^{2}x,$$

where

$$\mathrm{sech}x=\frac{1}{\cosh x}.$$

Once this structure is clear, use the MathsGenie bite-sized lessons and Question Bank practice to turn each identity into a method you can recall under exam pressure.

Hyperbolic functions worked examples

Solving a hyperbolic equation

Solve

$$\cosh x=\frac{5}{4}.$$

Start from the exponential definition:

$$\frac{e^x+e^{-x}}{2}=\frac{5}{4}.$$

Multiply by $2e^x$:

$$e^{2x}+1=\frac{5}{2}{e}^x.$$

Let $u=e^x$, where $u>0$:

$$2u^2-5u+2=0.$$

Factorise:

$$(2u-1)(u-2)=0.$$

Therefore

$$e^x=\frac{1}{2}or{e}^x=2.$$

Taking natural logarithms gives

$$x=-\ln 2orx=\ln 2.$$

The pair of answers makes sense because $\cosh x$ is even.

Differentiating a composite function

Differentiate

$$y=\tanh (3x).$$

Use

$$\frac{d}{dx}(\tanh x)={\mathrm{sech}}^{2}x$$

and apply the chain rule:

$$\frac{dy}{dx}=3{\mathrm{sech}}^2(3x).$$

A common mark is lost by writing only ${\mathrm{sech}}^2(3x)$. The factor of $3$ comes from differentiating the inner function.

Using an inverse hyperbolic function

Solve

$$\sinh x=2.$$

Apply the inverse function:

$$x=\mathrm{arsinh}(2).$$

Using

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

we obtain

$$x=\ln \left(2+\sqrt{5}\right).$$

Unlike $\cosh x$, the function $\sinh x$ is one-to-one, so there is only one real answer.

Inverse hyperbolic functions and domains

Inverse notation becomes easier once you connect it to graphs. A function has an inverse only where it is one-to-one. Since $\sinh x$ is strictly increasing across the real numbers, $\mathrm{arsinh}x$ accepts every real input. Its logarithmic form is

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

The function $\cosh x$ is not one-to-one on its full domain because it is even. Its inverse is therefore defined using the restricted branch $x\ge 0$. For $x\ge 1$,

$$\mathrm{arcosh}x=\ln \left(x+\sqrt{x^2-1}\right).$$

Similarly, because the range of $\tanh x$ is $(-1,1)$, its inverse accepts inputs only within that interval:

$$\mathrm{artanh}x=\frac{1}{2}\ln \left(\frac{1+x}{1-x}\right),-1<x<1.$$

These restrictions are part of the method, not an optional final check. If an equation leads to $\mathrm{arcosh}\left(\frac{1}{2}\right)$ or $\mathrm{artanh}(2)$, there is no real solution. In proof questions, state the relevant domain before using a logarithmic form. In numerical questions, check whether the calculator is set to return an exact expression or decimal approximation, and follow the requested accuracy carefully.

A reliable exam sequence is to isolate the hyperbolic function, check its possible range, apply the inverse, and then simplify. This prevents polished algebra from hiding an invalid answer.

Differentiation and integration rules

The most important derivatives are

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

and

$$\frac{d}{dx}(\tanh x)={\mathrm{sech}}^{2}x.$$

The first two rules are unusual because differentiating $\cosh x$ does not introduce a negative sign. This follows from their exponential definitions.

For integration,

$$\int \sinh (kx)dx=\frac{1}{k}\cosh (kx)+C,$$

and

$$\int \cosh (kx)dx=\frac{1}{k}\sinh (kx)+C.$$

Alex improves by checking each integral through differentiation. It takes a few seconds and catches missing factors before they become lost marks.

Hyperbolic functions exam questions

Try these without looking back at the rules:

1. Solve $\sinh x=3$, giving the answer in logarithmic form.
2. Differentiate $x\cosh (2x)$.
3. Find $\int \sinh (4x)dx$.
4. Show from the exponential definitions that ${\cosh }^{2}x-{\sinh }^{2}x=1$.
5. Explain why $\cosh x=-1$ has no real solution.

Answers:

1. $x=\ln (3+\sqrt{10})$.
2. $\cosh (2x)+2x\sinh (2x)$.
3. $\frac{1}{4}\cosh (4x)+C$.
4. Expand both squared definitions and simplify.
5. The real range of $\cosh x$ is $[1,\infty )$.

After topic practice, use the Edexcel Further Maths papers and mark schemes to see how these techniques appear within longer core pure questions. Students following another board can start from the MathsGenie revision homepage and select their course.

Common mistakes

Copying trigonometric signs

Do not replace ${\cos }^{2}x+{\sin }^{2}x=1$ with a hyperbolic identity containing the same sign. The correct identity is ${\cosh }^{2}x-{\sinh }^{2}x=1$.

Missing a second solution

If $\cosh x=k$ with $k>1$, expect two real solutions because $\cosh x$ is even. By contrast, $\sinh x=k$ has one real solution.

Confusing reciprocals and inverses

The expression $\mathrm{sech}x$ means $1/\cosh x$. The expression $\mathrm{arcosh}x$, sometimes written ${\cosh }^{-1}x$, means the inverse hyperbolic cosine. They are not the same operation.

Forgetting chain-rule factors

For example,

$$\frac{d}{dx}\sinh (5x)=5\cosh (5x),$$

not simply $\cosh (5x)$.

Ignoring domains

The logarithmic forms of inverse functions have restrictions. In particular, real $\mathrm{arcosh}x$ requires $x\ge 1$, while real $\mathrm{artanh}x$ requires $-1<x<1$.

Turn understanding into exam technique

Alex's turning point is not memorising a longer formula list. It is following the same sequence each time: identify the function, check its range, choose an identity or exponential definition, solve carefully, then test whether every answer is valid. Knowing that sequence matters, but practising it is what makes it available in an exam.

Your turn. Build that routine with the MathsGenie Lessons, then use the MathsGenie Further Maths Question Bank to practise hyperbolic functions exam questions until the method feels familiar.