Pa5 - Calculus of sinh, cosh and tanh | OCR A-Level Further Mathematics B (MEI)

The exponential definitions make hyperbolic differentiation and integration surprisingly direct. You derive the derivatives, use the chain and product rules, and recognise the corresponding antiderivatives, including a logarithm when you integrate tanh. Differentiating your answers will catch both missing scale factors and familiar trigonometric sign slips.

Differentiating sinh and cosh

Can you find the exact area under $y=\sinh x+\cosh x+\frac1{\cosh^2x}$ from $x=0$ to $x=\ln2$ ?

It looks like three new integration problems. By the third part, each term will have a simple antiderivative and the area will reduce to a fraction. First, derive the differentiation rules rather than guessing from ordinary trigonometry.

You already know

$\sinh x=\frac12(e^x-e^{-x})$ and $\cosh x=\frac12(e^x+e^{-x})$ . The chain rule gives $\frac{d}{dx}e^{-x}=-e^{-x}$ ; the product rule is $(uv)'=u'v+uv'$ .

Differentiate the definitions term by term:

\begin{aligned} \frac{d}{dx}\sinh x&=\frac12(e^x+e^{-x})=\cosh x&&\text{two minus signs cancel}\\ \frac{d}{dx}\cosh x&=\frac12(e^x-e^{-x})=\sinh x&&\text{differentiate the second exponential} \end{aligned}

The second derivative of each function is the function itself: $(\sinh x)''=\sinh x$ and $(\cosh x)''=\cosh x$ . Ordinary sine and cosine instead return with a minus sign. A tiny sign change produces very different curves.

Sinh and cosh derivatives

For a differentiable real function $u=u(x)$ :

\frac{d}{dx}\sinh u=u'\cosh u,\qquad \frac{d}{dx}\cosh u=u'\sinh u

There is no extra minus sign when differentiating hyperbolic cosine.

Not in the formula booklet: learn it.

When a hyperbolic function is multiplied by another function, use the product rule outside and the chain rule inside.

Worked example 1: product and chain rules together

Question. Differentiate $y=x\cosh(2x)$ .

Plan. Differentiate each factor in turn, including the inner derivative $2$ .

\begin{aligned} \frac{d}{dx}\cosh(2x)&=2\sinh(2x)&&\text{chain rule}\\ \frac{dy}{dx}&=1\cosh(2x)+x\big(2\sinh(2x)\big)&&\text{product rule}\\ &=\cosh(2x)+2x\sinh(2x)&&\text{simplify} \end{aligned}

Why this step? Differentiating only $\cosh(2x)$ would treat the factor $x$ as a constant. The product rule supplies the additional $\cosh(2x)$ term.

Answer. $\dfrac{dy}{dx}=\cosh(2x)+2x\sinh(2x)$ .

Check. At $x=0$ , the gradient is $1$ . Near zero, $\cosh(2x)$ is close to $1$ , so $x\cosh(2x)$ is close to $x$ , agreeing with that gradient.

The next example changes only cosine to sine. Follow the same product-rule structure and notice that no sign reversal appears.

Worked example 2: changing the hyperbolic factor

Question. Differentiate $y=x\sinh(2x)$ .

Plan. Use the product rule and differentiate the inner $2x$ by the chain rule.

\begin{aligned} \frac{d}{dx}\sinh(2x)&=2\cosh(2x)&&\text{chain rule}\\ \frac{dy}{dx}&=\sinh(2x)+2x\cosh(2x)&&\text{product rule} \end{aligned}

Why this step? Hyperbolic sine differentiates to hyperbolic cosine, while hyperbolic cosine differentiates back to hyperbolic sine. The sign difference from ordinary cosine has already been accounted for in the derivation.

Answer. $\dfrac{dy}{dx}=\sinh(2x)+2x\cosh(2x)$ .

Check. The derivative is $0$ at $x=0$ . The original function is even, since it is the product of two odd functions, so a horizontal tangent at the origin is consistent.

Writing the derivative of each factor before assembling the product helps keep the two rules separate.

Watch out

Wrong: $\frac{d}{dx}\cosh(3x)=-3\sinh(3x)$ , copied from cosine. Right: $\frac{d}{dx}\cosh(3x)=3\sinh(3x)$ . Derive it from the exponentials if the sign is uncertain.

The next question changes the power of the factor outside the hyperbolic function, so that factor's derivative changes too.

The derivative of tanh

Differentiating the ratio $\tanh x=\sinh x/\cosh x$ brings the fundamental identity into the calculation.

Apply the quotient rule and simplify its numerator:

\begin{aligned} \frac{d}{dx}\tanh x &=\frac{\cosh x\cosh x-\sinh x\sinh x}{\cosh^2x}&&\text{quotient rule}\\ &=\frac{1}{\cosh^2x}&&\cosh^2x-\sinh^2x=1 \end{aligned}

The derivative is positive everywhere, so the graph is increasing. At zero it is $1$ ; far from zero, $\cosh x$ is large and the derivative is small. That explains the flattening near the two horizontal asymptotes.

[DIAGRAM: asset_name: Tanh and its positive derivative; asset_slug: tanh_and_derivative; caption: Tanh increases everywhere, with its largest slope 1 at the origin and slopes tending to zero in both tails; description: Two aligned coordinate panels show tanh x in blue with dashed asymptotes y equals plus or minus 1, and its derivative 1 over cosh squared x in sage. The derivative panel peaks at 0,1 and stays above the x-axis. A tangent line at the origin in the first panel has gradient 1. Matching vertical guides at x equals minus 2,0,2 connect the idea of flatter tails to smaller derivative values.]
Tanh increases everywhere, with its largest slope 1 at the origin and slopes tending to zero in both tails

Tanh derivative

For a differentiable real function $u=u(x)$ :

\frac{d}{dx}\tanh u=\frac{u'}{\cosh^2u}

The reciprocal square $1/\cosh^2u$ is also written $\operatorname{sech}^2u$ , where $\operatorname{sech}u=1/\cosh u$ .

Not in the formula booklet: learn it.

Dividing the fundamental identity by $\cosh^2u$ also gives $1/\cosh^2u=1-\tanh^2u$ . This is a derived equivalent form, not a separate identity to memorise; it is not printed in the formula booklet.

Worked example 3: a scaled argument

Question. Differentiate $y=\tanh(2x)$ .

Plan. Use the tanh derivative and multiply by the derivative of $2x$ .

\begin{aligned} u&=2x,\qquad u'=2&&\text{identify the inner function}\\ \frac{dy}{dx}&=\frac{2}{\cosh^2(2x)}&&\text{chain rule} \end{aligned}

Why this step? The derivative of the inner function multiplies the reciprocal square; it does not replace the argument inside $\cosh$ .

Answer. $\dfrac{dy}{dx}=\dfrac{2}{\cosh^2(2x)}$ .

Check. At $x=0$ , the slope is $2$ . Replacing $x$ by $2x$ compresses the tanh graph horizontally and doubles its initial slope.

The next example adds one outer square. Differentiate that square before using the derivative from Worked example 3.

Worked example 4: a squared hyperbolic tangent

Question. Differentiate $y=\tanh^2(2x)$ .

Plan. Regard the function as the square of $\tanh(2x)$ , then apply the chain rule twice.

\begin{aligned} v&=\tanh(2x),\qquad y=v^2&&\text{separate the outer square}\\ \frac{dy}{dx}&=2v\frac{dv}{dx}&&\text{differentiate the square}\\ &=2\tanh(2x)\frac{2}{\cosh^2(2x)}&&\text{differentiate }v\\ &=\frac{4\tanh(2x)}{\cosh^2(2x)}&&\text{simplify} \end{aligned}

Why this step? $\tanh^2(2x)$ has a squared output. It is not $\tanh(4x)$ , whose input has been doubled.

Answer. $\dfrac{dy}{dx}=\dfrac{4\tanh(2x)}{\cosh^2(2x)}$ .

Check. The function is even and has minimum zero at the origin. Its derivative is odd and zero there, with the expected signs on either side.

Equivalent derivative forms are welcome when they are algebraically consistent. Keep powers and arguments visibly different.

Watch out

Wrong: $\frac{d}{dx}\tanh x=1+\tanh^2x$ , copied from ordinary tangent. Right: $\frac{d}{dx}\tanh x=1-\tanh^2x=1/\cosh^2x$ . The minus sign follows from the hyperbolic identity.

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Pa5 - Calculus of sinh, cosh and tanh

Differentiating sinh and cosh

The derivative of tanh

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