CP 8.2 - Hyperbolic derivatives and integrals | Edexcel A-Level Further Mathematics

Differentiate and integrate hyperbolic functions, combining their standard results with the chain, product and quotient rules (CP 8.2). You will see why both fundamental derivatives have positive signs and recognise integrals whose inner derivative is already present.

Derivatives and the chain rule

Why is the derivative of $\cosh x$ positive $\sinh x$ , when differentiating $\cos x$ produces a minus sign? The exponential definition answers the question before any rule needs memorising.

Differentiate its two terms:

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

The minus sign comes from differentiating $\mathrm e^{-x}$ ; it turns the half-sum into the half-difference. Similarly, differentiating the half-difference gives the half-sum, so $(\sinh x)'=\cosh x$ .

You already know

From A-level Mathematics, the chain rule multiplies by the derivative of the inner function; product and quotient rules handle combinations. From the previous lesson, $\cosh^2x-\sinh^2x=1$ , $\tanh x=\sinh x/\cosh x$ , and $\operatorname{sech}x=1/\cosh x$ .

The quotient rule now gives the third derivative:

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

It is positive for every real $x$ , agreeing with the increasing graph of $\tanh$ .

Standard hyperbolic derivatives

For real $x$ ,

\frac{\mathrm d}{\mathrm dx}\sinh x=\cosh x, \qquad \frac{\mathrm d}{\mathrm dx}\cosh x=\sinh x, \qquad \frac{\mathrm d}{\mathrm dx}\tanh x=\operatorname{sech}^2x

In the formula booklet.

For a differentiable inner function $u=u(x)$ , multiply each corresponding derivative by $u'$ . For instance, $\frac{\mathrm d}{\mathrm dx}\tanh u=u'\operatorname{sech}^2u$ . The chain rule is assumed knowledge; the booklet's displayed standard derivatives have argument $x$ , so you supply the inner derivative.

Worked example 1: a stretched argument

Question. Differentiate $y=\sinh3x$ with respect to $x$ .

Plan. Differentiate the outer hyperbolic function, then multiply by the derivative of $3x$ .

\begin{aligned} u&=3x,\qquad \frac{\mathrm du}{\mathrm dx}=3 && \text{inner function} \\ \frac{\mathrm dy}{\mathrm dx}&=\cosh u\times3 && \text{chain rule} \\ &=3\cosh3x && \text{return to }x \end{aligned}

Why this step? A change in $x$ changes the input $3x$ three times as fast, so the derivative acquires a factor of $3$ .

Answer. $\mathrm dy/\mathrm dx=3\cosh3x$ .

Check. Differentiating $(\mathrm e^{3x}-\mathrm e^{-3x})/2$ gives $3(\mathrm e^{3x}+\mathrm e^{-3x})/2$ , the same expression.

Change only the outer function from $\sinh$ to $\cosh$ . The inner factor stays the same, and there is still no extra minus sign.

Worked example 2: a hyperbolic cosine

Question. Differentiate $y=\cosh3x$ .

Plan. Use the derivative of $\cosh$ and retain the chain factor $3$ .

\begin{aligned} \frac{\mathrm dy}{\mathrm dx}&=\sinh3x\times\frac{\mathrm d}{\mathrm dx}(3x) && \text{chain rule} \\ &=3\sinh3x && \text{inner derivative} \end{aligned}

Why this step? The derivative of hyperbolic cosine is positive hyperbolic sine; importing the circular cosine rule would reverse the result.

Answer. $\mathrm dy/\mathrm dx=3\sinh3x$ .

Check. At $x=0$ , the derivative is zero, consistent with the minimum of $\cosh3x$ .

A sign error would also make the $\cosh$ curve decrease for positive inputs, contradicting its graph.

Watch out

Wrong: $(\cosh3x)'=-3\sinh3x$ . Right: $3\sinh3x$ . Keep the hyperbolic rules distinct from the circular trigonometric rules.

Use the same inner-derivative step for $\tanh$ .

Products, powers and quotients

A hyperbolic function inside a product or quotient obeys the same differentiation rules as every other differentiable function; identify the outer structure before you start.

For $x\sinh^2x$ , the outer structure is a product. Inside its second factor is a square, so that derivative needs the chain rule as well: $(\sinh^2x)'=2\sinh x\cosh x$ . For a denominator such as $\sqrt{x+1}$ , writing its reciprocal as $(x+1)^{-1/2}$ can make the product rule shorter than the quotient rule.

Worked example 3: a product containing a square

Question. Differentiate $y=x\sinh^2x$ .

Plan. Apply the product rule, differentiating the squared hyperbolic factor by the chain rule.

\begin{aligned} \frac{\mathrm dy}{\mathrm dx} &=1\times\sinh^2x+x\frac{\mathrm d}{\mathrm dx}(\sinh^2x) && \text{product rule} \\ &=\sinh^2x+2x\sinh x\cosh x && \text{chain rule for the square} \end{aligned}

Why this step? Both factors depend on $x$ , so differentiating just $\sinh^2x$ would omit the first product-rule term.

Answer. $\mathrm dy/\mathrm dx=\sinh^2x+2x\sinh x\cosh x$ .

Check. At $x=0$ , the expression gives zero. Near zero, $x\sinh^2x$ behaves like $x^3$ , so a horizontal tangent there is sensible.

Change the second factor to a reciprocal square, keeping the factor $x$ . Its negative exponent creates a minus sign through the power rule.

Worked example 4: a reciprocal hyperbolic square

Question. Differentiate $y=x/\cosh^2x$ .

Plan. Write $y=x(\cosh x)^{-2}$ , then apply the product and chain rules.

\begin{aligned} \frac{\mathrm dy}{\mathrm dx} &=(\cosh x)^{-2}-2x(\cosh x)^{-3}\sinh x && \text{product and chain rules} \\ &=\operatorname{sech}^2x-2x\operatorname{sech}^2x\tanh x && \text{reciprocal notation} \end{aligned}

Why this step? The negative exponent belongs to the outer power, so it contributes $-2$ ; differentiating $\cosh x$ itself still gives positive $\sinh x$ .

Answer. $\mathrm dy/\mathrm dx=\operatorname{sech}^2x-2x\operatorname{sech}^2x\tanh x$ .

Check. Direct quotient differentiation gives $(\cosh^2x-2x\cosh x\sinh x)/\cosh^4x$ , which simplifies to the same result.

For a square-root denominator, exactly the same negative-power approach works. The next pair keeps that denominator fixed while changing only the numerator function.

Worked example 5: an algebraic denominator

Question. Differentiate $y=\cosh2x/\sqrt{x+1}$ , for $x>-1$ .

Plan. Write the quotient as a product with $(x+1)^{-1/2}$ , then differentiate both factors.

\begin{aligned} y&=\cosh2x\,(x+1)^{-1/2} && \text{negative power} \\ \frac{\mathrm dy}{\mathrm dx} &=2\sinh2x\,(x+1)^{-1/2} -\frac12\cosh2x\,(x+1)^{-3/2} && \text{product rule} \end{aligned}

Why this step? Converting the denominator makes its derivative one power-rule line and keeps the two contributions visible.

Answer. $\mathrm dy/\mathrm dx=\frac{2\sinh2x}{\sqrt{x+1}}-\frac{\cosh2x}{2(x+1)^{3/2}}$ , for $x>-1$ .

Check. At $x=0$ , the answer is $-1/2$ : the numerator has zero derivative there while the increasing denominator makes the quotient decrease.

Replace $\cosh2x$ by $\sinh2x$ . All denominator work stays unchanged; only the numerator and its derivative exchange roles.

Worked example 6: changing the numerator function

Question. Differentiate $y=\sinh2x/\sqrt{x+1}$ , for $x>-1$ .

Plan. Use the same product structure and the derivative of $\sinh2x$ .

\begin{aligned} \frac{\mathrm dy}{\mathrm dx} &=2\cosh2x\,(x+1)^{-1/2} -\frac12\sinh2x\,(x+1)^{-3/2} && \text{product and chain rules} \\ &=\frac{2\cosh2x}{\sqrt{x+1}}-\frac{\sinh2x}{2(x+1)^{3/2}} && \text{fraction form} \end{aligned}

Why this step? The second product-rule term still contains the original numerator, $\sinh2x$ , because only the algebraic factor is differentiated in that term.

Answer. $\mathrm dy/\mathrm dx=\frac{2\cosh2x}{\sqrt{x+1}}-\frac{\sinh2x}{2(x+1)^{3/2}}$ , for $x>-1$ .

Check. At zero the derivative is $2$ , agreeing with the local behaviour $\sinh2x/\sqrt{x+1}\approx2x$ .

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account

Last Accessed

Core Pure Mathematics

Further Pure Mathematics 1

Further Pure Mathematics 2

Further Statistics 1

Further Statistics 2

Further Mechanics 1

Further Mechanics 2

Decision Mathematics 1

Decision Mathematics 2

CP 8.2 - Hyperbolic derivatives and integrals

Derivatives and the chain rule

Products, powers and quotients

Continue this lesson.

Learn faster with Nova