Problem 3 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(D_{x} y\). $$ y=5 \sinh ^{2} x $$

Step-by-Step Solution

Answer

The derivative of \(y = 5 \sinh^2 x\) is \(5 \cdot \sinh(2x)\).

Step 1: Understanding the Expression

The given exercise asks us to differentiate the function \(y = 5 \sinh^2 x\) with respect to \(x\). This involves using the chain rule to differentiate the hyperbolic sine function squared.

Step 2: Apply the Chain Rule

The chain rule states that the derivative of a function \(u^n\) is \(n \cdot u^{n-1} \cdot \frac{du}{dx}\). Here, let \(u = \sinh x\), so \(y = 5u^2\). The derivative of \(u^2\) is \(2u\). Therefore, differentiate \(5u^2\) with respect to \(u\) to get \(10u\).

Step 3: Derivative of \(u\)

The derivative of \(u = \sinh x\) with respect to \(x\) is \(\cosh x\). This is because the derivative of \(\sinh x\) is \(\cosh x\).

Step 4: Combine the Results

Using the results from Step 2 and Step 3, apply the chain rule: \[D_{x} y = 10u \cdot \cosh x = 10(\sinh x) \cdot \cosh x.\]

Step 5: Simplify the Expression

The expression \(10(\sinh x) \cdot \cosh x\) can be left in this form, or you can recognize that \(\sinh(2x) = 2\sinh x \cosh x\), using the identity to express it as \(\frac{5}{2} \sinh(2x)\). Thus the derivative can also be written as:\[D_{x} y = 5 \cdot \sinh(2x)\]

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus for differentiating composite functions. When dealing with a function composed of another function, like when you have a function of another function, you apply the chain rule. In simple terms, if you have a function that can be written as one function inside another, such as \( y = f(g(x)) \), then the derivative of \( y \) with respect to \( x \) is found by multiplying the derivative of the outer function evaluated at the inner function by the derivative of the inner function itself. The formula is:

• \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \)

In the context of our exercise, where \( y = 5 \sinh^2 x \), we identify \( g(x) = \sinh x \) and \( f(u) = 5u^2 \), with \( u = g(x) \). Applying the chain rule helps us express the derivative in terms of \( u \), then further simplify using the derivative of \( g(x) \). This step is crucial when dealing specifically with functions that aren't in their simplest form, enabling more complex differentiation.

Hyperbolic Functions

Hyperbolic functions, such as \( \sinh x \) and \( \cosh x \), are analogues of trigonometric functions but for hyperbolic geometry. These functions are very useful in calculus and complex analysis.

• \( \sinh x \) is the hyperbolic sine function defined as \( \sinh x = \frac{e^x - e^{-x}}{2} \).
• \( \cosh x \) is the hyperbolic cosine function defined as \( \cosh x = \frac{e^x + e^{-x}}{2} \).

For our problem, \( \sinh x \) is squared, giving 5 times the square of the hyperbolic sine. One key property of hyperbolic functions is their differentiability. Here, we know that the derivative of \( \sinh x \) is \( \cosh x \). This is analogous to how the derivative of \( \sin x \) is \( \cos x \) in trigonometric functions. These derivatives help us easily work through complex derivatives involving hyperbolic functions.

Derivative Identities

Derivative identities simplify our work by providing known results, which save time and effort in computation. In the case of hyperbolic functions, certain identities make differentiation smoother and sometimes even allow simplification of the results. For instance:

• \( \frac{d}{dx}[\sinh x] = \cosh x \)
• \( \frac{d}{dx}[\cosh x] = \sinh x \)

A particularly useful identity is \( \sinh(2x) = 2\sinh x \cosh x \), which allows us to reframe multiple terms down to more manageable expressions. In our exercise, this identity helps us to simplify \( 10 \sinh x \cosh x \) to \( 5\sinh(2x) \). Recognizing and applying these identities aids in both simplifying expressions and avoiding complexity in differentiation, ultimately leading to a cleaner, more elegant result.