4.07d - Differentiating and integrating hyperbolic functions | OCR A-Level Further Mathematics A

Differentiate and integrate hyperbolic functions by connecting their exponential definitions to familiar calculus rules. You will handle composite functions and products, find exact integrals, and use derivatives and the ranges of the functions to locate a curve's minimum.

Derivatives of sine and cosine hyperbolic

What is the least possible value of $\cosh^2x-4\sinh x$ , and at which input does it occur? At $x=0$ its value is $1$ , but that does not make it the minimum: the two terms change at different rates. Differentiation will tell you where their effects balance.

You already know

$\sinh x=(e^x-e^{-x})/2$ , $\cosh x=(e^x+e^{-x})/2$ , and $\cosh^2x-\sinh^2x=1$ . From A-level Maths, $\frac{d}{dx}e^{kx}=ke^{kx}$ ; the chain rule multiplies by the derivative of the inner function.

Differentiate each exponential separately. The two minus signs in the derivative of $-e^{-x}$ cancel:

\begin{aligned} \frac{d}{dx}\sinh x &= \frac12(e^x+e^{-x})=\cosh x \\ \frac{d}{dx}\cosh x &= \frac12(e^x-e^{-x})=\sinh x \end{aligned}

Unlike ordinary sine and cosine, these derivatives swap without a minus sign. In particular, each returns to itself after two differentiations. Both graphs solve the equation $y''=y$ : two quite different shapes can obey the same differential equation because their starting values differ.

Differentiating sine and cosine hyperbolic

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

Not in the formula booklet: learn it.

Read the entire argument before differentiating. The notation $\cosh 2x$ means $\cosh(2x)$ , whereas $\cosh^2x$ means $(\cosh x)^2$ .

Worked example 1: linear arguments

Question. Given that $f(x)=3\sinh 2x-\cosh 2x$ , find $f'(x)$ and $f''(x)$ .

Plan. Swap sine and cosine hyperbolic and multiply by $2$ at each differentiation.

\begin{aligned} f'(x)&=6\cosh 2x-2\sinh 2x && \text{chain rule} \\ f''(x)&=12\sinh 2x-4\cosh 2x && \text{chain rule again} \\ &=4f(x) && \text{factor out }4 \end{aligned}

Why this step? Each differentiation contributes a factor of $2$ , so two differentiations contribute $2^2$ , with no sign reversal.

Answer. $f'(x)=6\cosh 2x-2\sinh 2x$ , $f''(x)=12\sinh 2x-4\cosh 2x$ .

Check. $f(x)=e^{2x}-2e^{-2x}$ , whose derivatives are $2e^{2x}+4e^{-2x}$ and $4e^{2x}-8e^{-2x}$ , agreeing with the hyperbolic answers.

The second example changes only the inner argument, adding a constant. The factor from the chain rule stays the same because the derivative of the constant is zero.

Worked example 2: a shifted argument

Question. Given that $g(x)=3\sinh(2x+1)-\cosh(2x+1)$ , find $g'(x)$ and $g''(x)$ .

Plan. Differentiate the inner function $2x+1$ , then apply the same derivative pair.

\begin{aligned} \frac{d}{dx}(2x+1)&=2 && \text{inner derivative} \\ g'(x)&=6\cosh(2x+1)-2\sinh(2x+1) && \text{chain rule} \\ g''(x)&=12\sinh(2x+1)-4\cosh(2x+1) && \text{differentiate again} \\ &=4g(x) && \text{compare with }g \end{aligned}

Why this step? The $1$ stays inside the hyperbolic argument; it disappears only when finding the multiplying factor $u'$ .

Answer. $g'(x)=6\cosh(2x+1)-2\sinh(2x+1)$ , $g''(x)=4g(x)$ .

Check. At $x=-\frac12$ , the argument is zero, so $g=-1$ , $g'=6$ and $g''=-4=4g$ .

The usual chain, product and quotient rules continue to work; only the function derivatives have changed.

Watch out

Wrong: $\frac{d}{dx}\cosh 3x=-3\sinh 3x$ . That imports the minus sign from ordinary cosine. Right: $\frac{d}{dx}\cosh 3x=3\sinh 3x$ .

Use the inner derivative again, with a different scale.

The tangent derivative and composite functions

The derivative of $\tanh x$ comes from its definition as a quotient, so there is no need to guess a sign from ordinary tangent.

Using the quotient rule and the fundamental identity,

\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}=1-\tanh^2x && \text{divide the identity by }\cosh^2x \end{aligned}

Since $\cosh x\ge1$ , the derivative is positive everywhere and equals $1$ only at $x=0$ . Far from the origin it tends to zero, matching the flattening graph.

Differentiating tangent hyperbolic

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

\frac{d}{dx}\tanh u=\frac{u'}{\cosh^2u}=u'(1-\tanh^2u)

The denominator is non-zero for every real $u$ .
Not in the formula booklet: learn it.

Either form is useful. Use the quotient form when the answer contains $\cosh$ ; use $1-\tanh^2u$ when the question asks for tangent only.

Worked example 3: differentiating tangent hyperbolic

Question. Given that $y=\tanh 3x$ , find $\frac{dy}{dx}$ , expressing your answer in terms of $y$ .

Plan. Apply the chain rule and replace $\tanh 3x$ by $y$ .

\begin{aligned} \frac{dy}{dx}&=3(1-\tanh^2 3x) && \text{inner derivative }3 \\ &=3(1-y^2) && y=\tanh 3x \end{aligned}

Why this step? Replacing the whole function by $y$ expresses the rate of change using the output itself, as requested.

Answer. $\frac{dy}{dx}=3(1-y^2)$ .

Check. At $x=0$ , $y=0$ and the slope is $3$ , three times the slope of $\tanh x$ at zero.

Now square the same function. This adds one outer operation, so one more chain-rule factor is needed.

Worked example 4: an outer square

Question. Differentiate $y=\tanh^2 3x$ .

Plan. Differentiate the outer square first, then differentiate $\tanh 3x$ .

\begin{aligned} \frac{dy}{dx}&=2\tanh 3x\times\frac{d}{dx}(\tanh 3x) && \text{outer square} \\ &=2\tanh 3x\times\frac{3}{\cosh^2 3x} && \text{inner chain rule} \\ &=\frac{6\tanh 3x}{\cosh^2 3x} && \text{collect factors} \end{aligned}

Why this step? $\tanh^2 3x$ has three layers: square, tangent hyperbolic, then $3x$ . Every layer contributes a factor.

Answer. $\frac{dy}{dx}=6\tanh 3x/\cosh^2 3x$ .

Check. The function is even, so its derivative should be odd. The numerator is odd and the denominator is even, as required.

For a product, apply the product rule before simplifying. For example,

\frac{d}{dx}(x\tanh x)=\tanh x+\frac{x}{\cosh^2x}

The product rule is in your assumed A-level Maths knowledge; it is not given in the formula booklet. Its two terms are both needed.

Watch out

Wrong: $\frac{d}{dx}\tanh^2 3x=3/\cosh^2 3x$ . This differentiates only the inner tangent and forgets the square. Right: multiply that derivative by $2\tanh 3x$ .

Check the three layers with a new inner scale.

Integrating hyperbolic functions

Integration reverses the derivatives, so a constant inner scale must be divided out rather than multiplied in.

The tangent integral needs one extra observation. Since $\cosh x>0$ , the logarithmic derivative gives

\frac{d}{dx}\ln(\cosh x)=\frac{\sinh x}{\cosh x}=\tanh x

There is no absolute-value issue here: $\cosh x$ is already positive for every real input. The general logarithmic integration rule is in the formula booklet; the hyperbolic results below are not.

Continue this lesson.

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

Create free account

Last Accessed

Pure Core

Statistics

Mechanics

Discrete Mathematics

Additional Pure Mathematics

4.07d - Differentiating and integrating hyperbolic functions

Derivatives of sine and cosine hyperbolic

The tangent derivative and composite functions

Integrating hyperbolic functions

Continue this lesson.

Learn faster with Nova