Problem 25 - Thomas Calculus

Question

In Exercises \(19-30,\) find \(d y\) $$ y=\sin (5 \sqrt{x}) $$

Step-by-Step Solution

Answer

The differential \( dy \) is \( \cos(5\sqrt{x}) \cdot \frac{5}{2\sqrt{x}} \cdot dx \).

Step 1: Identify the Function

The given function is \( y = \sin(5\sqrt{x}) \). It is a composite function involving both a sine function and a square root function. To find the differential \( dy \), we'll need to find the derivative \( y' \) first.

Step 2: Apply Chain Rule

Recall that the chain rule states that if \( y = f(g(x)) \), then \( y' = f'(g(x)) \cdot g'(x) \). Here, \( f(u) = \sin(u) \) and \( g(x) = 5\sqrt{x} \). We'll first differentiate \( f(u) \) with respect to \( u \) and \( g(x) \) with respect to \( x \).

Step 3: Differentiate the Outer Function

The derivative of \( f(u) = \sin(u) \) with respect to \( u \) is \( f'(u) = \cos(u) \). Thus, we have \( \cos(5\sqrt{x}) \) for the outer function.

Step 4: Differentiate the Inner Function

The function \( g(x) = 5\sqrt{x} \) is actually \( 5x^{1/2} \). The derivative \( g'(x) \) is \( \frac{d}{dx}[5x^{1/2}] = \frac{5}{2}x^{-1/2} = \frac{5}{2\sqrt{x}} \).

Step 5: Combine Using the Chain Rule

Apply the chain rule: \( y' = \cos(5\sqrt{x}) \cdot \frac{5}{2\sqrt{x}} \). This gives us the derivative of \( y \) with respect to \( x \).

Step 6: Derive \( dy \) in Terms of \( dx \)

Since \( y' = \frac{dy}{dx} = \cos(5\sqrt{x}) \cdot \frac{5}{2\sqrt{x}} \), it follows that \[ dy = \cos(5\sqrt{x}) \cdot \frac{5}{2\sqrt{x}} \cdot dx \].

Key Concepts

Chain Rule

The chain rule is an essential tool in differential calculus, particularly when dealing with composite functions. It allows us to compute the derivative of a function composed of two or more functions. Suppose you have a function in the form of

• \( y = f(g(x)) \)

To find the derivative \( y' \), the chain rule tells us to multiply the derivative of the outer function \( f \) evaluated at \( g(x) \) by the derivative of the inner function \( g(x) \). In mathematical terms, the chain rule is expressed as:

• \( y' = f'(g(x)) \, \cdot g'(x) \)

This rule is fundamental when functions are nested, ensuring that every component of the function is properly accounted for in the differentiation process.
In our exercise, the function is \( y = \sin(5\sqrt{x}) \). Here, \( \sin \, \, y = f(u) \) is the outer function, while \( u = g(x) = 5\sqrt{x} \) is the inner function.Understanding the chain rule makes handling composite functions manageable.

Composite Function

Composite functions arise when one function is nested inside another. In mathematical terms, a composite function \( h(x) \) can be expressed as:

• \( h(x) = f(g(x)) \)

This involves applying the function \( g \) to \( x \) and then the function \( f \) to the result of \( g(x) \).
To illustrate, consider our example \( y = \sin(5\sqrt{x}) \). This is a classic composite function:

• The inner function is \( g(x) = 5\sqrt{x} \),
• and the outer function is \( f(u) = \sin(u) \).

Recognizing composite functions is crucial for applying the chain rule effectively in differentiation.
Composite functions may involve more than two functions, hence requiring multiple applications of the chain rule.By understanding composite functions, you can break down complex expressions into manageable parts.

Derivative

A derivative is a measure of how a function changes as its input changes, capturing the idea of the rate of change or slope of a function at any given point.
Derivatives are foundational in calculus, allowing us to understand and predict the behavior of functions.
In finding the derivative of a composite function, like our example \( y = \sin(5\sqrt{x}) \), we apply the chain rule.
First, identify the derivatives of the individual components:

• The derivative of the outer function \( \sin(u) \) is \( \cos(u) \),
• and the derivative of the inner function \( 5\sqrt{x} \) is \( \frac{5}{2\sqrt{x}} \).

Applying the chain rule, we get
\[ y' = \cos(5\sqrt{x}) \, \cdot \frac{5}{2\sqrt{x}} \],
which represents the derivative of the entire function.
Calculating derivatives correctly allows us to determine the behavior and characteristics of curves.

Differentiation Steps

Differentiation involves several systematic steps for finding the derivative of a function accurately. For composite functions, these steps can be summarized as:
**1. Identify all components of the function:**
- Recognize the outer function \( f \) and inner function \( g \).
For example, in \( y = \sin(5\sqrt{x}) \), \( \sin(u) \) is outer, \( 5\sqrt{x} \) is inner.
**2. Differentiate the outer function (\( f \)):**
- Compute \( f'(u) \), the derivative of \( f \).
Here, it's \( \cos(u) \).
**3. Differentiate the inner function (\( g \)):**
- Compute \( g'(x) \), the derivative of \( g(x) \).
For our example, \( \frac{5}{2\sqrt{x}} \) is the derivative.
**4. Apply the chain rule:**
- Multiply the derivatives: \( f'(g(x)) \, \cdot \, g'(x) \).
These steps build on one another to ensure you apply the chain rule properly.
Ultimately, these steps result in accurately solving the derivative for composite functions.