Differentiating hyperbolic functionsEdexcel International A Level Further Maths: Revision notes

Section 1

The basic derivatives

From the definitions $\sinh x=\frac{e^x-e^{-x}}{2}$ and $\cosh x=\frac{e^x+e^{-x}}{2}$ , differentiating term by term gives $\frac{d}{dx}\sinh x=\cosh x,\qquad\frac{d}{dx}\cosh x=\sinh x,\qquad\frac{d}{dx}\tanh x=\operatorname{sech}^2x.$ For $\tanh x=\frac{\sinh x}{\cosh x}$ the quotient rule gives $\frac{\cosh^2x-\sinh^2x}{\cosh^2x}=\frac{1}{\cosh^2x}$ . Note that $\operatorname{sech}x=\frac{1}{\cosh x}$ and $\operatorname{sech}^2x=1-\tanh^2x$ . Unlike the trigonometric functions, there are no minus signs: $\cosh x$ differentiates to $\sinh x$ , not $-\sinh x$ .

Key termshyperbolic functionsech

Common mistake

Writing $\frac{d}{dx}\cosh x=-\sinh x$ by analogy with $\cos x$ . The result is $+\sinh x$ .

Exam tip

Check a result with $e^x$ : both $\sinh x$ and $\cosh x$ differentiate to the other one.

Section 2

The chain rule

If the argument is not just $x$ , multiply by its derivative: $\frac{d}{dx}\sinh(kx)=k\cosh(kx),\quad\frac{d}{dx}\cosh(kx)=k\sinh(kx),\quad\frac{d}{dx}\tanh(kx)=k\operatorname{sech}^2(kx).$ For a power, $\frac{d}{dx}\sinh^2x=2\sinh x\cosh x=\sinh2x$ . For a general inner function, $\frac{d}{dx}\sinh\left(g(x)\right)=g'(x)\cosh\left(g(x)\right)$ . Example: $\frac{d}{dx}\tanh3x=3\operatorname{sech}^23x$ .

Key termschain rule

Common mistake

Forgetting the factor $k$ , or writing $3\operatorname{sech}^2x$ and losing the $3x$ inside.

Exam tip

For $\sinh^2x$ the outer function is the square: $2\sinh x$ times the derivative of $\sinh x$ .

Section 3

Products and quotients

Combine the basic results with the product rule $(uv)'=u'v+uv'$ and the quotient rule $\left(\frac uv\right)'=\frac{u'v-uv'}{v^2}$ . Product: $\frac{d}{dx}\left(x\sinh^2x\right)=\sinh^2x+2x\sinh x\cosh x$ . Quotient: $\frac{d}{dx}\left(\frac{\cosh2x}{x+1}\right)=\frac{(x+1)\cdot2\sinh2x-\cosh2x}{(x+1)^2}$ . Factorise where it helps: $\sinh^2x+2x\sinh x\cosh x=\sinh x\left(\sinh x+2x\cosh x\right)$ .

Key termsproduct rulequotient rule

Common mistake

In the quotient rule, subtracting in the wrong order. The numerator is $u'v-uv'$ .

Section 4

Stationary points and tangents

Set $\frac{dy}{dx}=0$ and solve. Equations such as $\sinh x=2$ or $\tanh x=\frac34$ are solved with the logarithmic forms, or by writing everything with $e^x$ and solving a quadratic in $e^x$ . Example: $y=\cosh x-2x$ . Then $\sinh x=2$ , so $e^x-e^{-x}=4$ , giving $e^{2x}-4e^x-1=0$ and $e^x=2+\sqrt5$ , $x=\ln(2+\sqrt5)$ . The second derivative decides the nature. For $y=4\cosh x-3\sinh x$ , $\frac{d^2y}{dx^2}=y$ , so a stationary point with $y>0$ is a minimum. Use $\cosh^2x-\sinh^2x=1$ to move between $\sinh$ , $\cosh$ and $\tanh$ . Remember $e^x>0$ , so reject any negative root for $e^x$ .

Key termsstationary pointsecond derivative

Common mistake

Keeping the root $e^x=-1$ or $e^x=2-\sqrt5$ . Since $e^x>0$ it must be rejected.

Exam tip

Express $\sinh$ and $\cosh$ using $e^x$ when the equation is not a simple $\sinh x=k$ .

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Differentiating hyperbolic functions

The hyperbolic functions are defined by $\sinh x=\frac{e^x-e^{-x}}{2}$ , $\cosh x=\frac{e^x+e^{-x}}{2}$ and $\tanh x=\frac{\sinh x}{\cosh x}$ , and $\operatorname{sech}x=\frac{1}{\cosh x}$ .
Use the quotient rule to show that $\frac{d}{dx}\left(\tanh x\right)=\operatorname{sech}^2x$ .2 marks
The curve $C$ has equation $y=x\sinh^2x$ .
Show that $C$ has exactly one stationary point.2 marks
The function $f$ is defined by $f(x)=\frac{\cosh2x}{x+1}$ for $x>-1$ .
Find $f'(x)$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).