Problem 14 - Thomas Calculus

Question

In Exercises \(13-24,\) find the derivative of \(y\) with respect to the appropriate variable. $$ y=\frac{1}{2} \sinh (2 x+1) $$

Step-by-Step Solution

Answer

The derivative is \( \cosh(2x + 1) \).

Step 1: Identify the Derivative Formula for Hyperbolic Sine

To find the derivative of the hyperbolic sine function, identify the derivative formula. The derivative of \( \sinh(u) \) with respect to \( u \) is \( \cosh(u) \). This formula will be used as part of the chain rule.

Step 2: Apply the Chain Rule

Since the argument of the hyperbolic sine function \( \sinh \) is \( (2x+1) \), use the chain rule. Let \( u = 2x + 1 \), so \( y = \frac{1}{2} \sinh(u) \). The chain rule states that \( \frac{d}{dx} \sinh(u) = \cosh(u) \cdot \frac{du}{dx} \).

Step 3: Differentiate the Inner Function

Differentiate the inner function \( u = 2x + 1 \) with respect to \( x \). The derivative \( \frac{du}{dx} \) is \( 2 \).

Step 4: Differentiate the Entire Expression

Combine the derivative of \( \sinh(u) \) with the derivative of \( u \). The final derivative is \( \frac{d}{dx} y = \frac{1}{2} \cdot \cosh(2x + 1) \cdot 2 \).

Step 5: Simplify the Expression

Simplify the expression by multiplying the constants: \( \frac{d}{dx} y = \cosh(2x + 1) \). This is your final derivative.

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used to find the derivative of a composite function. A composite function is one in which a function is applied inside another function. For the function given, \( y = \frac{1}{2} \sinh(2x + 1) \), the hyperbolic sine function is applied to \( 2x + 1 \), a linear function.

When utilizing the chain rule, you first need to identify both the "outer" function and the "inner" function:

• The outer function is \( \sinh(u) \) where \( u = 2x + 1 \).
• The inner function is \( u = 2x + 1 \).

According to the chain rule, if you want to find the derivative of a composite function \( f(g(x)) \), it is expressed as \( f'(g(x)) \cdot g'(x) \). This involves differentiating the outer function first and then multiplying it by the derivative of the inner function. In this example, this technique efficiently helps in differentiating the hyperbolic sine function.

Derivative

Derivatives are a core part of calculus. They measure how a function changes as its input changes. Specifically, they provide the rate at which a quantity changes with respect to a change in another quantity. For the function \( y = \frac{1}{2} \sinh(2x + 1) \), finding the derivative helps in understanding how \( y \) changes as \( x \) changes.

To find the derivative of our function with the chain rule, we:

• Differentiate \( \sinh(u) \), which is our outer function. The derivative of \( \sinh(u) \) is \( \cosh(u) \).
• Apply the chain rule by multiplying \( \cosh(u) \) with the derivative of the inner function \( u = 2x + 1 \).

The inner function has a simple linear form, \( u = 2x + 1 \), and its derivative \( \frac{du}{dx} \) is \( 2 \). By putting these together, we derive the whole function, demonstrating how the derivative analyzes the rate of change in specific mathematical terms.

Mathematical Differentiation

Mathematical differentiation refers to the process of finding a derivative. It's a tool used to compute how a function's output changes in response to changes in its input. Differentiation is widely applied in various fields, helping to model and solve problems involving rates of change for different quantities.

In our current example, differentiation goes through the following stages:

• Identifying the functions involved, here being \( \frac{1}{2}\sinh(u) \), where \( u = 2x + 1 \).
• Applying the derivative of the hyperbolic sine function, \( \cosh(u) \), as part of the process.
• Combining the derivative of our independent variable, \( u \), to apply the chain rule effectively.

After distinguishing these parts, the final step involves simplification. In this case, you multiply the constants and simplify the final expression to obtain \( \frac{d}{dx} y = \cosh(2x + 1) \). This shows how differentiation can break down complex expressions into manageable parts.