HF2: Derivatives and integrals of hyperbolic functions

The differentiation and integration of hyperbolic functions allow us to understand how these functions change and accumulate values which teaches us the essential tools for analysing complex mathematical and real-world systems. Keep reading to learn about these functions and put your new knowledge to the test with some exercises.

See Hyperbolic functions for a list of definitions.

The derivatives of the hyperbolic functions may be found using their
definitions.

For \(\sinh,\) let \(a\) be a constant, then:
\[\begin{align*}
\frac{d}{dx}\sinh\left(ax\right) & =\frac{d}{dx}\left(\frac{e^{ax}-e^{-ax}}{2}\right)\\
& =\frac{ae^{ax}+ae^{-ax}}{2}\\
& =a\cosh\left(ax\right).
\end{align*}\]
For cosh,
\[\begin{align*}
\frac{d}{dx}\cosh\left(ax\right) & =\frac{d}{dx}\left(\frac{e^{ax}+e^{-ax}}{2}\right)\\
& =\frac{ae^{ax}-ae^{-ax}}{2}\\
& =a\sinh\left(ax\right).
\end{align*}\]

The quotient, product and chain rules can be applied to functions
involving hyperbolic functions. For example, using the quotient rule,
\[\begin{align*}
\frac{d}{dx}\tanh\left(ax\right) & =\frac{d}{dx}\left(\frac{\sinh\left(ax\right)}{\cosh\left(ax\right)}\right)\\
& =\frac{d}{dx}\left(\frac{e^{ax}-e^{-ax}}{e^{ax}+e^{-ax}}\right)\\
& =\frac{\left(e^{ax}+e^{-ax}\right)\frac{d}{dx}\left(e^{ax}-e^{-ax}\right)-\left(e^{ax}-e^{-ax}\right)\frac{d}{dx}\left(e^{ax}+e^{-ax}\right)}{\left(e^{ax}+e^{-ax}\right)^{2}}\\
& =\frac{\left(e^{ax}+e^{-ax}\right)a\left(e^{ax}+e^{-ax}\right)-a\left(e^{ax}-e^{-ax}\right)\left(e^{ax}-e^{-ax}\right)}{\left(e^{ax}+e^{-ax}\right)^{2}}\\
& =\frac{a\left(e^{ax}+e^{-ax}\right)^{2}-a\left(e^{ax}-e^{-ax}\right)^{2}}{\left(e^{ax}+e^{-ax}\right)^{2}}\\
& =a-a\frac{\left(e^{ax}-e^{-ax}\right)^{2}}{\left(e^{ax}+e^{-ax}\right)^{2}}\\
& =a\left(1-\tanh^{2}\left(ax\right)\right)\\
& =a\text{sech$^{2}\left(ax\right).$ }
\end{align*}\]

In summary,
\[\begin{align*}
\frac{d}{dx}\sinh\left(ax\right) & =a\cosh\left(ax\right)\\
\frac{d}{dx}\cosh\left(ax\right) & =a\sinh\left(ax\right)\\
\frac{d}{dx}\tanh\left(ax\right) & =a\text{sech$^{2}$ $\left(ax\right)$ .}
\end{align*}\]

Example 1

Find the derivative, with respect to \(x\), of \(\cosh\left(x^{2}+3x\right).\)

Solution

Let \(u=x^{2}+3x\) then \(du/dx=2x+3\) and by the chain rule,
\[\begin{align*}
\frac{d}{dx}\cosh\left(x^{2}+3x\right) & =\frac{d}{du}\cosh\left(u\right)\frac{du}{dx}\\
& =\sinh\left(u\right)\left(2x+3\right)\\
& =\left(2x+3\right)\sinh\left(x^{2}+3x\right).
\end{align*}\]
Hence the derivative, with respect to \(x,\) of \(\cosh\left(x^{2}+3x\right)\)
is \(\left(2x+3\right)\sinh\left(x^{2}+3x\right)\).

Example 2

Find the approximate slope of the tangent to \(y=\sinh\left(4x\right)\)
at \(x=0.5\) to three decimal places.

Solution

Differentiating,
\[\begin{align*}
\frac{dy}{dx} & =4\cosh\left(4x\right).
\end{align*}\]
At \(x=0.5,\)
\[\begin{align*}
\left.\frac{dy}{dx}\right|_{x=0.5} & =4\cosh\left(4\left(0.5\right)\right)\\
& =4\cosh\left(2\right)\\
& =4\left(\frac{e^{2}+e^{-2}}{2}\right)\\
& =2\left(e^{2}+\left(\frac{1}{e^{2}}\right)\right)\\
& \approx15.049\,.
\end{align*}\]
The slope of the tangent is approximately \(15.049\) at \(x=0.5\).

Integrals of hyperbolic functions

Let \(a\) be a constant. Using the definitions,
\[\begin{align*}
\int\sinh\left(ax\right)dx & =\int\left(\frac{e^{ax}-e^{-ax}}{2}\right)dx\\
& =\frac{\frac{1}{a}e^{ax}+\frac{1}{a}e^{-ax}}{2}+c,\quad c\in\mathbb{R}\\
& =\frac{1}{a}\cosh\left(ax\right)+c.
\end{align*}\]
Similarly,
\[\begin{align*}
\int\cosh\left(ax\right)dx & =\int\left(\frac{e^{ax}+e^{-ax}}{2}\right)dx\\
& =\frac{\frac{1}{a}e^{ax}-\frac{1}{a}e^{-ax}}{2}+c,\quad c\in\mathbb{R}\\
& =\frac{1}{a}\sinh\left(ax\right)+c.
\end{align*}\]
For the integral of \(\tanh,\) we use integration by substitution.
Let \(u=\cosh\left(ax\right),\) then \(du/dx=a\,\sinh\left(x\right)\)
and
\[\begin{align*}
\int\tanh\left(ax\right)dx & =\int\frac{\sinh\left(ax\right)}{\cosh\left(ax\right)}dx\\
& =\frac{1}{a}\int\frac{1}{\cosh\left(ax\right)}\frac{du}{dx}dx\\
& =\frac{1}{a}\int\frac{1}{u}du\\
& =\frac{1}{a}\ln\left(\cosh\left(ax\right)\right)+c,\quad c\in\mathbb{R}.
\end{align*}\]

Summarising
\[\begin{align*}
\int\sinh\left(ax\right)dx & =\frac{1}{a}\cosh\left(ax\right)+c\\
\int\cosh\left(ax\right)dx & =\frac{1}{a}\sinh\left(ax\right)+c\\
\int\tanh\left(ax\right)dx & =\frac{1}{a}\ln\left(\cosh\left(ax\right)\right)+c.
\end{align*}\]

Example 3

Find \(\int\cosh\left(3x\right)dx.\)

Solution

\[\begin{align*}
\int\cosh\left(3x\right) & =\frac{1}{3}\sinh\left(3x\right)+c
\end{align*}\]
where \(c\) is a constant.

Example 4

Find the value of \(\int_{0}^{\ln\left(2\right)}\sinh\left(x\right)dx.\)

Solution

\[\begin{align*}
\int_{0}^{\ln\left(2\right)}\sinh\left(x\right)dx & =\left[\cosh\left(x\right)\right]_{x=0}^{x=\ln\left(2\right)}\\
& =\cosh\left(\ln\left(2\right)\right)-\cosh\left(0\right)\\
& =\frac{1}{2}\left(e^{\ln\left(2\right)}+e^{-\ln\left(2\right)}\right)-\frac{1}{2}\left(e^{0}+e^{-0}\right)\\
& =\frac{1}{2}\left(e^{\ln\left(2\right)}+e^{\ln\left(\frac{1}{2}\right)}\right)-1\\
& =\frac{1}{2}\left(2+\frac{1}{2}\right)-1\\
& =\frac{1}{2}\left(\frac{5}{2}\right)-1\\
& =\frac{1}{4}.
\end{align*}\]

Example 5

Find \(\int\left(12x^{3}-2\right)\tanh\left(3x^{4}-2x\right)dx.\)

Solution

We use the substitution method. Let
\[\begin{align*}
u & =3x^{4}-2x
\end{align*}\]
then
\[\begin{align*}
\frac{du}{dx} & =12x^{3}-2.
\end{align*}\]
Now, the integral may be written
\[\begin{align*}
\int\left(12x^{3}-2\right)\tanh\left(3x^{4}-2x\right)dx & =\int\frac{du}{dx}\tanh\left(u\right)dx\\
& =\int\tanh\left(u\right)du\\
& =\ln\left(\cosh\left(u\right)\right)+c'\,,\ \text{where $c'\ \text{is a constant.}$ }\\
& =\ln\left(\cosh\left(3x^{4}-2x\right)\right)+c\,,\ \text{where $c\ \text{is a constant.}$ }
\end{align*}\]
Hence
\[\begin{align*}
\int\left(12x^{3}-2\right)\tanh\left(3x^{4}-2x\right)dx & =\ln\left(\cosh\left(3x^{4}-2x\right)\right)+c\,,\ \text{where $c\ \text{is a constant.}$ }
\end{align*}\]

Exercises

1. Find the derivative, with respect to \(x,\) of
   \(\quad\)a) \(y=6\cosh\left(x/3\right)\)
   \(\quad\)b) \(y=\frac{1}{2}\sinh\left(2x+1\right)\)
2. Evaluate:
   \(\quad\)a) \(\int\cosh\left(3x\right)dx\)
   \(\quad\)b) \(\int_{1}^{2}\frac{\cosh\left(\ln\left(t\right)\right)}{t}dt.\)

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