2.3 Differentiation

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📘 OBJECTIVE 1 — Differentiating Hyperbolic and Inverse Functions

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What Are Hyperbolic Functions?

You already know the ordinary trigonometric functions like $\sin x$ and $\cos x$. Hyperbolic functions look similar but are defined using the number $e$ (Euler's number, approximately 2.718). They appear in physics, engineering, and advanced maths.

The three main hyperbolic functions are:

$\sinh x = \frac{e^x - e^{-x}}{2}, \qquad \cosh x = \frac{e^x + e^{-x}}{2}, \qquad \tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$

• $\sinh x$ is read as "shine x" or "sinch x"
• $\cosh x$ is read as "cosh x"
• $\tanh x$ is read as "than x" or "tanch x"

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Derivatives of the Hyperbolic Functions

To differentiate these, you can use their $e^x$ definitions. Here are the results you must know:

Why does $\dfrac{d}{dx}(\sinh x) = \cosh x$?

$\frac{d}{dx}\!\left(\frac{e^x - e^{-x}}{2}\right) = \frac{e^x + e^{-x}}{2} = \cosh x \checkmark$

Why does $\dfrac{d}{dx}(\cosh x) = \sinh x$?

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

Why does $\dfrac{d}{dx}(\tanh x) = \text{sech}^2 x$?

Use the quotient rule on $\tanh x = \dfrac{\sinh x}{\cosh x}$:

$\frac{d}{dx}(\tanh x) = \frac{\cosh x \cdot \cosh x - \sinh x \cdot \sinh x}{\cosh^2 x} = \frac{\cosh^2 x - \sinh^2 x}{\cosh^2 x}$

Using the identity $\cosh^2 x - \sinh^2 x = 1$:

$$= \frac{1}{\cosh^2 x} = \text{sech}^2 x \checkmark$$

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✏️ Worked Example 1

Differentiate $y = \sinh(3x^2)$.

Use the chain rule: differentiate the outside function, then multiply by the derivative of the inside.

$\frac{dy}{dx} = \cosh(3x^2) \cdot 6x = 6x\cosh(3x^2)$