Question

Differentiating Hyperbolic Functions Evaluate the following derivatives: a. \(\frac{d}{d x}\left(\sinh \left(x^{2}\right)\right)\) b. \(\frac{d}{d x}(\cosh x)^{2}\)

Short answer

a. \\(2x \cosh(x^2)\\), b. \\(\sinh(2x)\\).

Step-by-step solution

Differentiate using the Chain Rule for a

To differentiate \(rac{d}{dx} \sinh(x^2)\), recognize that it's a composition of functions. We use the chain rule: if \(u = x^2\), then \(\frac{d}{dx} \sinh(u) = \cosh(u) \cdot \frac{du}{dx}\). Since \(\frac{du}{dx} = 2x\), the derivative is \(\cosh(x^2) \cdot 2x\).

Differentiate the Hyperbolic Function for a

Now apply the values from the chain rule: the derivative is \(2x \cosh(x^2)\).

Differentiate using the Product Rule for b

For \(\frac{d}{dx}((\cosh x)^2)\), express it as \(\cosh x \cdot \cosh x\), which will require the product rule: \(u \cdot v' + v \cdot u'\). Here, both \(u\) and \(v\) are \(\cosh x\) with \(u' = \sinh x\).

Apply the Product Rule for b

Using the product rule: \(\cosh x \cdot \sinh x + \cosh x \cdot \sinh x\). This simplifies to \(2 \sinh x \cosh x\), which can be written using the identity for the hyperbolic sine of double angle as \(\sinh(2x)\).

Write Final Derivatives

For part (a), the derivative is \(2x \cosh(x^2)\) and for part (b), the derivative is \(\sinh(2x)\).

Key concepts

Hyperbolic Functions

Hyperbolic functions, similar to their trigonometric counterparts, describe relationships in hyperbolic geometry. The most common hyperbolic functions are the hyperbolic sine and hyperbolic cosine, denoted as \( \sinh x \) and \( \cosh x \) respectively. These functions are defined as follows:
\[ \sinh x = \frac{e^x - e^{-x}}{2} \]
\[ \cosh x = \frac{e^x + e^{-x}}{2} \]
The intriguing part of hyperbolic functions is their resemblance to circular functions but with different geometric properties. They are widely used in various fields, including physics, engineering, and mathematics.

• \( \sinh x \) represents the hyperbolic sine.
• \( \cosh x \) gives us the hyperbolic cosine.

These functions follow identities similar to trigonometric identities; for example, \( \cosh^2 x - \sinh^2 x = 1 \), much like \( \cos^2 x + \sin^2 x = 1 \). This similarity makes them useful in solving certain calculus problems.

Chain Rule

The Chain Rule in calculus is an essential tool for differentiating compositions of functions. When a function is a composition, like \( \sinh(x^2) \), the Chain Rule helps in finding the derivative.
The rule states that for two functions \( f(g(x)) \), the derivative \( \frac{d}{dx} f(g(x)) \) is \( f'(g(x)) \cdot g'(x) \). The rule helps us decompose problems into smaller ones, making it easier to manage.

Here's a simple way to use it:

• Identify the inside function: Here, \( g(x) = x^2 \).
• Find the derivative of the outside function with respect to the inside function: \( \frac{d}{du} \sinh(u) = \cosh(u) \).
• Differentiate the inside function: \( \frac{d}{dx} x^2 = 2x \).

This process yields \( \cosh(x^2) \cdot 2x \), our derivative. The Chain Rule is especially beneficial when dealing with nested functions.

Product Rule

The Product Rule is another important technique in calculus, particularly when dealing with the differentiation of products of two separate functions. If \( y = u(x) \, v(x) \), the product rule states that:
\[ \frac{d}{dx}(u(x) \, v(x)) = u'(x) \, v(x) + u(x) \, v'(x) \]
This is useful when functions can’t be easily separated or simplified.

Let's apply this to \( (\cosh x)^2 \):

• Consider \( u(x) = \cosh x \) and \( v(x) = \cosh x \).
• Find \( u'(x) = \sinh x \) and \( v'(x) = \sinh x \).

The Product Rule gives us \( \cosh x \, \sinh x + \cosh x \, \sinh x = 2 \cosh x \, \sinh x \). A lovely outcome is that it simplifies further to \( \sinh(2x) \), thanks to the double angle identity for hyperbolic functions. Recognizing when to use this rule can greatly simplify your work.

Calculus

Calculus forms the backbone of mathematical analysis, dealing with change and motion. It is divided mainly into differential and integral calculus. Differentiation, the core topic here, involves finding how a function's output changes with respect to changes in input. This is essentially finding the derivative, which tells us the rate of change.

• Differential calculus focuses on finding the derivative of functions.
• Integral calculus concerns itself with the accumulation of quantities.

Using differentiation, we solve problems like optimization, motion, and many more real-world applications. By employing rules such as the Chain Rule and Product Rule, we can handle even more complex functions.
Calculus not only serves science and engineering but also industries like economics and medicine, providing essential insights into dynamic systems. Grasping these fundamental concepts can open doors to understanding the world’s mathematical underpinnings.