Question

Compute the derivative of the given function. $$f(x)=x^{2} \sin (5 x)$$

Short answer

The derivative is \(f'(x) = 2x\sin(5x) + 5x^2\cos(5x)\).

Step-by-step solution

Identify the Type of Function

The given function is a product of two functions: a polynomial \(x^2\) and a trigonometric function \(\sin(5x)\). We will use the product rule to find the derivative.

Recall the Product Rule

The product rule for derivatives states that if you have two functions \(u(x)\) and \(v(x)\), then the derivative of their product is \(u'(x)v(x) + u(x)v'(x)\).

Assign Functions

Let \(u(x) = x^2\) and \(v(x) = \sin(5x)\). We need to find \(u'(x)\) and \(v'(x)\).

Compute \(u'(x)\)

The derivative of \(u(x) = x^2\) is \(u'(x) = 2x\).

Compute \(v'(x)\)

First, apply the chain rule to \(v(x) = \sin(5x)\). The outer function is \(\sin(x)\), and the inner function is \(5x\). So, \(v'(x) = \cos(5x) \cdot 5 = 5\cos(5x)\).

Apply the Product Rule

Now that we have \(u'(x)=2x\) and \(v'(x)=5\cos(5x)\), we can use the product rule: \[\begin{align*}f'(x) &= u'(x)v(x) + u(x)v'(x) \&= (2x)(\sin(5x)) + (x^2)(5\cos(5x)) \&= 2x\sin(5x) + 5x^2\cos(5x).\end{align*}\]

Simplify the Expression

The expression for the derivative \(f'(x)\) is already simplified. Therefore, the derivative of the function \(f(x) = x^2 \sin(5x)\) is \(f'(x) = 2x\sin(5x) + 5x^2\cos(5x)\).

Key concepts

Derivative

Derivatives are a core concept in calculus that measure how a function changes as its input changes. Imagine you're on a road trip and you want to know how fast you're going at any given moment. Your speedometer gives you a reading, and this is essentially what a derivative does for functions. It tells us the rate at which a function is changing at any point. In the context of the exercise, we're finding the derivative of the function \( f(x) = x^2 \sin(5x) \) to understand how this combination of polynomial and trigonometric functions behaves as \( x \) varies.

If a function is represented as \( f(x) \), the derivative is denoted as \( f'(x) \) or \( \frac{d}{dx}f(x) \). The process involves determining how small changes to \( x \) impact \( f(x) \). With derivatives, we can solve real-world problems like finding maximum or minimum points, understanding dynamic systems, and predicting future behavior of systems.

Product Rule

The product rule is essential when dealing with derivatives of products of two or more functions. Instead of finding derivatives of each function separately and attempting to combine them, the product rule allows us to calculate accurately in one go.

Imagine you have two functions, \( u(x) \) and \( v(x) \). According to the product rule, if you need to find the derivative of their product \( u(x) \cdot v(x) \), the formula is given by:

• \( \frac{d}{dx}[u(x) \cdot v(x)] = u'(x)v(x) + u(x)v'(x) \)

Rather than simply multiplying derivatives, the product rule accounts for the interaction between the changing rates of the functions. In the given exercise, \( u(x) = x^2 \) and \( v(x) = \sin(5x) \), these apply directly to find \( f'(x) \). Without the product rule, we might miss important dynamic interactions in combined functions.

Chain Rule

The chain rule is a fundamental tool in calculus for finding the derivative of composite functions. Think of it as a technique that helps us "chain together" the derivatives of functions nested within each other. When you have a function inside another function, like \( \sin(5x) \), you need the chain rule to find its derivative.

To apply the chain rule, consider any function \( g(x) = f(h(x)) \). The derivative \( g'(x) \) can be found using:

• \( \frac{d}{dx}f(h(x)) = f'(h(x)) \cdot h'(x) \)

In our exercise, \( h(x) = 5x \) is inside the trigonometric function \( \sin(x) \). First, we differentiate the outer function "sine," and then multiply by the derivative of the inner linear function "5x." This results in \( \frac{d}{dx}\sin(5x) = \cos(5x) \cdot 5 = 5\cos(5x) \), which is crucial for finding \( v'(x) \).

Trigonometric Functions

Trigonometric functions such as \( \sin(x) \), \( \cos(x) \), and \( \tan(x) \) are vital in both mathematics and physics. They describe the relationships between angles and lengths in triangles but also model wave-like patterns in nature, such as sound and light waves.

In the context of derivatives, it's important to remember the basic derivatives of trigonometric functions:

• The derivative of \( \sin(x) \) is \( \cos(x) \)
• The derivative of \( \cos(x) \) is \(-\sin(x) \)
• The derivative of \( \tan(x) \) is \( \sec^2(x) \)

In our exercise, \( \sin(5x) \) is a trigonometric function, and its derivative involves recognizing its pattern with the chain rule. We multiply by the derivative of the inner function to correctly compute the full derivative. Understanding these patterns helps us solve complex calculus problems involving trigonometric relationships.