H2 - Differentiating and integrating hyperbolic functions | AQA A-Level Further Mathematics

Differentiating sinh gives cosh, and differentiating cosh gives sinh: there is no alternating minus sign. You will derive the rules for all six hyperbolic functions, reverse them in integration, include chain factors, and construct the less obvious logarithmic antiderivatives with their domains.

Derivatives and integrals of six hyperbolic functions

At $x=\ln2$ , is the gradient of $y=\cosh x$ equal to $3/4$ or $-3/4$ ? Here the two exponential values are 2 and $1/2$ ; differentiating their half-sum will settle the sign.

The negative exponent contributes a minus sign of its own:

\frac{\mathrm d}{\mathrm dx}\sinh x =\frac{e^x+e^{-x}}2=\cosh x \qquad \frac{\mathrm d}{\mathrm dx}\cosh x =\frac{e^x-e^{-x}}2=\sinh x

At the opening input, the second rule gives gradient $\sinh(\ln2)=(2-1/2)/2=3/4$ . Its positive sign agrees with the rising cosh graph there.

Reciprocal and quotient rules then give the remaining derivatives. For instance, $(1/\cosh x)'=-\sinh x/\cosh^2x=-\operatorname{sech}x\tanh x$ . Replacing cosh by sinh gives $(1/\sinh x)'=-\cosh x/\sinh^2x=-\operatorname{cosech}x\coth x$ , for $x\ne0$ .

Six derivatives

Function	Derivative
$\sinh x$	$\cosh x$
$\cosh x$	$\sinh x$
$\tanh x$	$\operatorname{sech}^2x$
$\operatorname{sech}x$	$-\operatorname{sech}x\tanh x$
$\operatorname{cosech}x$	$-\operatorname{cosech}x\coth x$
$\coth x$	$-\operatorname{cosech}^2x$

The last two rows hold for $x\ne0$ ; all other rows hold for every real $x$ .

The tanh derivative is in the formula booklet. The other five hyperbolic derivatives are not: learn or derive them.

The quotient calculations explain the opposite signs in the two squared derivatives.

Worked example 1: differentiating tanh

Question. Use the quotient rule to differentiate $y=\tanh x$ .

Plan. Write tanh as sinh divided by cosh, then use the fundamental identity.

\frac{\mathrm dy}{\mathrm dx} =\frac{\cosh x\cosh x-\sinh x\sinh x}{\cosh^2x} =\frac1{\cosh^2x}=\operatorname{sech}^2x

Answer. $\mathrm dy/\mathrm dx=\operatorname{sech}^2x$ , for real $x$ .

Check. The derivative is positive everywhere, agreeing with the strictly increasing graph of tanh. Direct differentiation of $(e^{2x}-1)/(e^{2x}+1)$ gives $4e^{2x}/(e^{2x}+1)^2$ , the same expression.

Now reverse the quotient. This reverses the order of the squared terms in the numerator.

Worked example 2: differentiating coth

Question. Use the quotient rule to differentiate $y=\coth x$ , where $x\ne0$ .

Plan. Write coth as cosh divided by sinh and keep the quotient-rule subtraction in order.

\frac{\mathrm dy}{\mathrm dx} =\frac{\sinh^2x-\cosh^2x}{\sinh^2x} =-\frac1{\sinh^2x}=-\operatorname{cosech}^2x

Answer. $\mathrm dy/\mathrm dx=-\operatorname{cosech}^2x$ , for $x\ne0$ .

Check. The reciprocal rule applied to $1/\tanh x$ gives $-\operatorname{sech}^2x/\tanh^2x=-1/\sinh^2x$ , independently confirming the minus sign.

Integration reverses these rules. Thus $\int\operatorname{sech}^2x\,\mathrm dx=\tanh x+C$ , $\int\operatorname{cosech}^2x\,\mathrm dx=-\coth x+C$ , and the two reciprocal-product derivatives integrate with a minus sign. Integrating the six functions themselves gives the following reference; the less immediate forms are explained in the final part.

Integrand	One antiderivative, plus $C$	Real interval
$\sinh x$	$\cosh x$	Any real interval
$\cosh x$	$\sinh x$	Any real interval
$\tanh x$	$\ln(\cosh x)$	Any real interval
$\coth x$	$\ln\lvert\sinh x\rvert$	An interval excluding zero
$\operatorname{sech}x$	$2\tan^{-1}(e^x)$	Any real interval
$\operatorname{cosech}x$	$\ln\lvert\tanh(x/2)\rvert$	An interval excluding zero

Here $\tan^{-1}$ means inverse tangent, not a reciprocal. Other correct antiderivatives can differ by a constant. Separate intervals on opposite sides of zero may have different constants.

Watch out

Wrong: $(\cosh x)'=-\sinh x$ . Unlike the circular cosine rule, $(\cosh x)'=\sinh x$ : differentiating $e^{-x}$ supplies the required subtraction within sinh itself.

Apply the reciprocal derivatives and the original domain together.

Chain factors and substitutions

An inner function changes the derivative by a chain factor. If $u=u(x)$ , then $(\sinh u)'=u'\cosh u$ and $(\cosh u)'=u'\sinh u$ . The same factor $u'$ multiplies each of the other four derivative rules.

For a linear input $kx+b$ , with $k\ne0$ , reversing this rule divides the antiderivative by $k$ . Do not divide by the entire input. For a nonlinear input, look for its derivative as another factor in the integrand.

Worked example 3: multiplying by the inner derivative

Question. Differentiate $y=\cosh(3x+1)$ .

Plan. Differentiate cosh first, then multiply by the derivative of $3x+1$ .

\frac{\mathrm dy}{\mathrm dx}=\sinh(3x+1)\times3=3\sinh(3x+1)

Answer. $3\sinh(3x+1)$ .

Check. Differentiating $(e^{3x+1}+e^{-3x-1})/2$ directly gives $(3e^{3x+1}-3e^{-3x-1})/2$ , the same result.

Keep the same function and reverse the operation. The factor now divides rather than multiplies.

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H2 - Differentiating and integrating hyperbolic functions

Derivatives and integrals of six hyperbolic functions

Chain factors and substitutions

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