Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(d y\). $$ y=\left(1+\sinh ^{3} 2 x\right)^{1 / 2} $$

Step-by-Step Solution

Answer

The derivative \( dy \) is \( 3 \sinh^2(2x) \cdot \cosh(2x) \cdot (1 + \sinh^3(2x))^{-1/2} \).

Step 1: Recognize the Composition of Functions

The given function is \( y = \left(1 + \sinh^3(2x)\right)^{1/2} \). Notice that it is a composition of the outer function \( u^{1/2} \) where \( u = 1 + \sinh^3(2x) \). We will use the chain rule to differentiate \( y \) with respect to \( x \).

Step 2: Differentiate the Outer Function

Differentiate the outer function \( u^{1/2} \) with respect to \( u \). The derivative is \( \frac{1}{2} u^{-1/2} \). Applying this to our function gives us \( \frac{1}{2} \left(1 + \sinh^3(2x)\right)^{-1/2} \cdot \frac{d}{dx}(u) \).

Step 3: Differentiate the Inner Function

Now, differentiate the inner function \( u = 1 + \sinh^3(2x) \) with respect to \( x \). Use the chain rule again: the derivative of \( \sinh^3(2x) \) is \( 3\sinh^2(2x) \cdot \cosh(2x) \cdot 2 \) (since the derivative of \( \sinh(kx) \) is \( k\cosh(kx) \)).

Step 4: Substitute and Simplify

Substitute the derivatives found in Steps 2 and 3 back into the chain rule setup:\[dy = \frac{1}{2} \left(1 + \sinh^3(2x)\right)^{-1/2} \cdot 3 \sinh^2(2x) \cdot \cosh(2x) \cdot 2 \]Simplifying, we have:\[dy = 3 \sinh^2(2x) \cdot \cosh(2x) \cdot \left(1 + \sinh^3(2x)\right)^{-1/2}\]

Key Concepts

Chain Rule

The chain rule is a fundamental concept in differential calculus, used when differentiating composite functions. Imagine a function wrapped inside another function. The chain rule breaks this down by differentiating the outer function and then multiplying by the derivative of the inner function. This process allows us to tackle complex functions step by step. When using the chain rule, identify:

• The outer function, which we differentiate first.
• The inner function, whose derivative we calculate next.

For example, in our problem, the outer function is \(u^{1/2}\), and the inner function is \(1 + \sinh^3(2x)\). By methodically differentiating these parts, we can successfully find the derivative of the entire function. Understanding and applying the chain rule is crucial for solving various problems that involve composed functions.

Hyperbolic Functions

Hyperbolic functions, like \(\sinh\) and \(\cosh\), are analogues of the trigonometric functions but are based on hyperbolas rather than circles. These functions are essential in calculus, particularly in differential equations and calculus involving areas such as physics and engineering. The key hyperbolic functions include:

• The hyperbolic sine, \(\sinh(x)\), defined as \[\sinh(x) = \frac{e^x - e^{-x}}{2}\]. \
• The hyperbolic cosine, \(\cosh(x)\), defined as \[\cosh(x) = \frac{e^x + e^{-x}}{2}\].\

In our example, we use \(\sinh^3(2x)\). The derivative of \(\sinh(x)\) is \(\cosh(x)\), which is key to applying the chain rule. Recognizing and understanding these functions helps in calculating derivatives correctly in problems involving hyperbolic terms.

Derivatives

Derivatives provide the rate of change of a function with respect to a variable. They are a cornerstone in calculus, portraying how a function changes at any point. Calculating derivatives involves various rules, and one crucial technique is the use of derivatives of hyperbolic functions. Knowing the derivative of \(\sinh(x)\) as \(\cosh(x)\) and \(\cosh(x)\) as \(\sinh(x)\), allows you to find derivatives smoothly. In our example expression, we derive \(\sinh^3(2x)\). Using the power rule and recognizing that \(\frac{d}{dx}\sinh(kx) = k\cosh(kx)\), permits the successful application of the chain rule. Furthermore, simplifying and substituting derivatives efficiently leads to finding the required differential \(dy\). Mastering derivatives not only solves homework problems but aids in comprehending broader calculus concepts, like function behaviors and optimizations.