Problem 9 - Calculus Single Variable

Question

Use the Product Rule to compute the derivative of the given expression with respect to \(x\). (In each of Exercises 7,8,14,16, and 18, do not avoid using the Product Rule by first expanding the product.) $$ x^{3} \cos (x) $$

Step-by-Step Solution

Answer

The derivative of \(x^3 \cos(x)\) is \(3x^2 \cos(x) - x^3 \sin(x)\).

Step 1: Identify Functions

The given expression is the product of two functions: the first is \( u(x) = x^3 \) and the second is \( v(x) = \cos(x) \). These will be differentiated using the product rule.

Step 2: Recall the Product Rule

The Product Rule states that if you have two functions \( u(x) \) and \( v(x) \), then the derivative of their product \( u(x) v(x) \) is given by \( (uv)' = u'v + uv' \). We'll apply this to \( x^3 \cos(x) \).

Step 3: Differentiate Each Function

Differentiate \( u(x) = x^3 \) to find \( u'(x) = 3x^2 \). Differentiate \( v(x) = \cos(x) \) to find \( v'(x) = -\sin(x) \).

Step 4: Apply the Product Rule

Substitute into the Product Rule formula: \((uv)' = u'v + uv'\). Therefore, \((x^3 \cdot \cos(x))' = (3x^2)(\cos(x)) + (x^3)(-\sin(x))\).

Step 5: Simplify the Expression

Simplify the expression obtained: \(3x^2 \cos(x) - x^3 \sin(x)\). This is the derivative of the expression \(x^3 \cos(x)\).

Key Concepts

Differentiation techniques

Differentiation techniques are essential tools in calculus that allow us to find the rate at which a function is changing at any given point. Among the various techniques available, the Product Rule is particularly important when dealing with the derivatives of products of two functions. In this context, we look at the expression involving products like \(x^3 \cos(x)\).

• The Product Rule formula is given by \((u \cdot v)' = u' \cdot v + u \cdot v'\), where \(u\) and \(v\) are functions of \(x\).
• This technique is essential when you can't or are instructed not to first expand the product before differentiating.
• It helps in breaking down complex expressions into simpler derivatives.

When approaching problems using the Product Rule, it's crucial to carefully identify each part of the expression before diving into differentiation. This makes it easier to plug into the formula and solve step by step.

Calculus

Calculus is a fundamental branch of mathematics that focuses on limits, functions, derivatives, integrals, and infinite series. One of its core concepts is the derivative, which measures how a function changes as its input changes.
Derivatives are vital for understanding the behavior of curves and for solving numerous practical problems involving rates of change. For instance:

• In physics, derivatives describe velocity as the rate of change of position with respect to time.
• In economics, they can represent the rate of change in cost or revenue concerning production levels.

By mastering calculus and its differentiation techniques, students can gain a deeper insight into how different variables interact dynamically in various fields. This kind of mathematical problem-solving is indispensable for creating mathematical models, predicting outcomes, and making informed decisions.

Trigonometric functions

Trigonometric functions like \(\sin(x)\), \(\cos(x)\), and \(\tan(x)\) are crucial in the study of calculus and especially in problems involving differentiation. These functions often arise in various practical and theoretical contexts such as physics, engineering, and even everyday problem-solving situations.
The differentiation of trigonometric functions is fundamental in calculus. For example:

• The derivative of \(\cos(x)\) is \(-\sin(x)\).
• Similarly, the derivative of \(\sin(x)\) is \(\cos(x)\).

Understanding these derivatives allows us to tackle more complex problems, as we saw with the derivative of the expression \(x^3 \cos(x)\). Especially when the Product Rule is involved, knowing these basic derivatives assists in breaking down and simplifying more intricate expressions.

Mathematical problem-solving

Mathematical problem-solving is about applying mathematical concepts and techniques to find solutions to complex problems. In the realm of calculus, this often involves breaking down a problem into manageable parts, applying relevant differentiation techniques, and simplifying the results.
Consider the derivative of \(x^3 \cos(x)\):

• First, recognize the structure of the expression and identify the functions involved.
• Next, utilize the Product Rule to differentiate the expression effectively.
• Finally, simplify the resulting derivative to its cleanest form: \(3x^2 \cos(x) - x^3 \sin(x)\).

By following these steps, a complex problem can be managed in a more straightforward, systematic manner. Such an approach not only helps in calculus but also in other areas of mathematics and science where logical problem-solving is required.