Problem 73 - Single Variable Calculus

Question

Prove that \(\frac{d}{d x} \cosh u=(\sinh u) \frac{d u}{d x} .\)

Step-by-Step Solution

Answer

In summary, we proved that \(\frac{d}{d x} \cosh u=(\sinh u) \cdot \frac{d u}{d x}\) by first defining the hyperbolic cosine function and finding its derivative with respect to \(u\), identifying the derivative as the hyperbolic sine function, and finally applying the chain rule. So, the short answer is: \(\frac{d}{d x} \cosh u=(\sinh u) \cdot \frac{d u}{d x}\).

Step 1: Define the hyperbolic cosine function

The hyperbolic cosine function can be defined as: \[ \cosh u = \frac{e^u + e^{-u}}{2} \]

Step 2: Find the derivative of the hyperbolic cosine function with respect to u

Using the definition of the hyperbolic cosine, we can find its derivative with respect to \(u\). We differentiate \(e^u\) and \(e^{-u}\) separately, and apply the chain rule when differentiating \(e^{-u}\): \[ \frac{d}{d u} \cosh u = \frac{d}{d u} \left( \frac{e^u + e^{-u}}{2} \right) = \frac{1}{2} \cdot \left( \frac{d e^u}{d u} + \frac{d e^{-u}}{d u} \right) \] \[ = \frac{1}{2} \cdot \left( e^u - e^{-u} \right) \]

Step 3: Identify the derivative as the hyperbolic sine function

From the result of step 2, we can see that the derivative of \(\cosh u\) with respect to \(u\) is equal to the hyperbolic sine function, since: \[ \sinh u = \frac{e^u - e^{-u}}{2} \] So, we can write the result of step 2 as: \[ \frac{d}{d u} \cosh u = \sinh u \]

Step 4: Apply the chain rule

Now, we will apply the chain rule to find the derivative of \(\cosh u\) with respect to \(x\). Recall that the chain rule states: \[ \frac{d}{d x} \cosh u = \frac{d}{d u} \cosh u \cdot \frac{d u}{d x} \] Substitute the result from step 3: \[ \frac{d}{d x} \cosh u = (\sinh u) \cdot \frac{d u}{d x} \] Therefore, we have proved that \(\frac{d}{d x} \cosh u=(\sinh u) \cdot \frac{d u}{d x}\).

Key Concepts

Chain Rule

The chain rule is a fundamental tool for differentiating composite functions. It allows us to differentiate functions that are composed of two or more simpler functions, connected in such a way that the output of one becomes the input of the next. Imagine a function written as \( f(u(x)) \), where \( u(x) \) itself is a function of \( x \). According to the chain rule, the derivative of \( f \) with respect to \( x \) is given by:\[ \frac{d}{dx} f(u) = \frac{df}{du} \cdot \frac{du}{dx} \]In simple terms, this means you first differentiate \( f \) with respect to \( u \), and then multiply by the derivative of \( u \) with respect to \( x \).

• It is a powerful method that simplifies the differentiation of complex chains of functions.
• You will encounter it often when working with various kinds of functions, including hyperbolic functions.

For example, in our exercise, we apply the chain rule to the derivative of \( \cosh u \) with respect to \( x \). By recognizing that \( u \) is itself a function of \( x \), we can write the derivative in terms of \( \sinh u \) and \( \frac{d u}{d x} \), illustrating the chain rule's utility in such scenarios.

Derivative of Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but based on hyperbolas rather than circles. Their derivatives have distinct properties that are useful in calculus. The two primary hyperbolic functions are the hyperbolic sine and cosine, denoted \( \sinh \) and \( \cosh \) respectively. They are defined mathematically as:- \( \sinh u = \frac{e^u - e^{-u}}{2} \)- \( \cosh u = \frac{e^u + e^{-u}}{2} \)Differentiating these functions requires us to understand their respective rates of change:

• The derivative of \( \cosh u \) with respect to \( u \) is \( \sinh u \).
• The derivative of \( \sinh u \) with respect to \( u \) is \( \cosh u \).

These relationships are central to proving identities or integrating hyperbolic functions within calculus problems. In our exercise, we specifically focus on differentiating \( \cosh u \) by using the definition of \( \cosh \) in terms of exponentials. This involves applying regular differentiation rules to \( e^u \) and \( e^{-u} \), confirming that \( \frac{d}{d u} \cosh u = \sinh u \). This step then plays a crucial role in the chain rule application.

Hyperbolic Sine

The hyperbolic sine function, symbolized as \( \sinh \), is an essential hyperbolic function with important properties that resemble yet differ from the conventional sine function in trigonometry. Defined as \( \sinh u = \frac{e^u - e^{-u}}{2} \), it plays a pivotal part in describing many real-world phenomena, including growth models and relativistic calculations. Key characteristics of \( \sinh u \) include:

• Its odd nature, meaning \( \sinh(-u) = -\sinh(u) \).
• Its derivative is \( \cosh u \), making it integral to differentiating hyperbolic functions.

In the context of the exercise solution, \( \sinh u \) represents the derivative of \( \cosh u \) with respect to \( u \). This relationship is essential when utilizing the chain rule, allowing us to express further derivatives, like \( \frac{d}{d x} \cosh u \), in a simplified form involving \( \sinh u \). Therefore, mastering \( \sinh u \) and its derivative properties helps understand hyperbolic function behavior and their applications in calculus.