﻿ How to Differentiate Hyperbolic Trigonometric Functions. Visual Explanation with color coded examples

How to Differentiate Hyperbolic Trigonometric Functions

Quick Overview

• $$\displaystyle \frac d {dx}\left(\sinh kx\right) = k\cosh kx$$
• $$\displaystyle \frac d {dx}\left(\cosh kx\right) = k\sinh kx$$
• $$\displaystyle \frac d {dx}\left(\tanh kx\right) = k\operatorname{sech}^2 kx$$
• $$\displaystyle \frac d {dx}\left(\coth kx\right) = -k\operatorname{csch}^2 kx$$
• $$\displaystyle \frac d {dx}\left(\operatorname{sech} kx\right) = -k\operatorname{sech} kx\tanh kx$$
• $$\displaystyle \frac d {dx}\left(\operatorname{csch} kx\right) = -k\operatorname{csch} kx\coth kx$$
• Notice that these derivatives are nearly identical to the "normal" trig derivatives. The only exception is the negative signs on the derivatives of the $$\cosh x$$ and $$\operatorname{sech} x$$.
• The trig functions are paired when it comes to differentiation: sinh and cosh, tanh and sech, coth and csch.
• This lesson assumes you are familiar with the $$\blue{Power\hspace{2mm}Rule}$$ , $$\blue{Product\hspace{2mm}Rule}$$, $$\blue{Quotient\hspace{2mm}Rule}$$ and $$\blue{Chain\hspace{2mm}Rule}$$.

Examples

Example 1

Find $$\displaystyle \frac d {dx}\left(\cosh(x^2+9)\right)$$.

Step 1

Use the chain rule to differentiate.

$$\frac d {dx}\left(\cosh(x^2+9)\right) = \sinh(x^2+9)\cdot \frac d {dx}\left(x^2 + 9\right) = \sinh(x^2+9)\cdot 2x = 2x\sinh(x^2+9)$$

$$\displaystyle \frac d {dx}\left(\cosh(x^2+9)\right) = 2x\sinh(x^2+9)$$

Example 2

Suppose $$f(x) = x^4\tanh 3x$$. Find $$f'(x)$$.

Step 1

Identify the factors in the function.

$$f(x) = \blue{x^4}\red{\tan 3x}$$

Step 2

Differentiate using the product rule .

\begin{align*}% f'(x) & = \blue{4x^3}\tanh 3x + x^4\cdot \red{3\operatorname{sech}^2 3x}\\ & = 4x^3\tanh 3x + 3x^4\sech^2 3x \end{align*}

Step 3

Simplify by factoring.

\begin{align*} f'(x) & = 4\blue{x^3}\tanh 3x + 3\blue{x^4}\sech^2 3x\\ & = \blue{x^3}\left(4\tanh 3x + 3x\sech^2 3x\right) \end{align*}

$$\displaystyle f'(x) = x^3\left(4\tanh 3x + 3x\sech^2 3x\right)$$.

Example 3

Find $$\displaystyle \frac d {dx}\left(\frac{\sinh 8x}{1 + \sech 8x}\right)$$.

Step 1

Differentiate using the quotient rule. The parts in $$\blue{blue}$$ are related to the numerator.

\begin{align*} \frac d {dx}\left(\frac{\sinh 8x}{1 + \sech 8x}\right) & = \frac{(1 + \sech 8x)\cdot\blue{8\cosh 8x} - \blue{\sinh 8x}\cdot (-8)\sech 8x\tanh 8x}{(1 + \sech 8x)^2} \end{align*}

Step 2

Simplify the numerator.

\begin{align*} \frac d {dx}\left(\frac{\sinh 8x}{1 + \sech 8x}\right) & = \frac{(1 + \sech 8x)\cdot \blue 8\cosh 8x - \sinh 8x\cdot \blue{(-8)}\sech 8x\tanh 8x}{(1 + \sech 8x)^2}\\[6pt] & = \frac{\blue 8(1 + \red{\sech 8x})\cdot \cosh 8x + \blue 8 \sinh 8x\cdot \red{\sech 8x}\tanh 8x}{(1 + \sech 8x)^2}\\[6pt] & = \frac{\blue 8\left(1 + \red{ \frac 1 {\cosh 8x}}\right)\cdot \cosh 8x + \blue 8\sinh 8x\cdot\red{\frac 1 {\cosh 8x}}\,\tanh 8x}{(1 + \sech 8x)^2}\\[6pt] & = \frac{\blue 8\left(\cosh 8x + 1\right) + \blue 8\cdot\frac{\sinh 8x}{\cosh 8x}\cdot\tanh 8x}{(1 + \sech 8x)^2}\\[6pt] & = \frac{\blue 8\left(\cosh 8x + 1\right) + \blue 8\cdot\tanh 8x\cdot\tanh 8x}{(1 + \sech 8x)^2}\\[6pt] & = \frac{\blue 8\left(\cosh 8x + 1 + \tanh^2 8x\right)}{(1 + \sech 8x)^2} \end{align*}

$$\displaystyle \frac d {dx}\left(\frac{\sinh 8x}{1 + \sech 8x}\right) = \frac{8\left(\tanh^2 8x + \cosh 8x + 1\right)}{(1 + \sech 8x)^2}$$

Example 4

Suppose $$f(x) = \coth^5 11x$$. Find $$f'(x)$$.

Step 1

Rewrite the function to emphasize that the cotangent is being raised to the fifth power.

$$f(x) = \left(\coth 11x\right)^5$$

Step 2

Differentiate using the chain rule.

\begin{align*} f'(x) & = 5\left(\coth 11x\right)^4\cdot \frac d {dx}\left(\coth 11x\right)\\[6pt] & = 5\left(\coth 11x\right)^4\cdot (-11\csch^2 11x)\\[6pt] & = -55\left(\coth 11x\right)^4\cdot \csch^2 11x \end{align*}

$$\displaystyle f'(x) = -55\left(\coth 11x\right)^4\cdot \csch^2 11x$$.