Question

Find \(D_{x} y\) using the rules of this section. $$ y=\left(x^{2}+2\right)\left(x^{3}+1\right) $$

Short answer

The derivative is \( D_x y = 5x^4 + 6x^2 + 2x \).

Step-by-step solution

Identify the Formula to Use

The expression given is a product of two functions. To differentiate this expression, we will use the Product Rule. The Product Rule states that if you have two functions \( u(x) \) and \( v(x) \), the derivative of their product \( u(x)v(x) \) is given by \( u'(x)v(x) + u(x)v'(x) \).

Define Functions and Derivatives

Let \( u(x) = x^2 + 2 \) and \( v(x) = x^3 + 1 \). Now, find the derivatives: - For \( u(x) = x^2 + 2 \), the derivative \( u'(x) = 2x \).- For \( v(x) = x^3 + 1 \), the derivative \( v'(x) = 3x^2 \).

Apply the Product Rule

Using the product rule, differentiate the function: \[ D_x y = u'(x)v(x) + u(x)v'(x) \] Substitute the expressions for the functions and their derivatives: \[ D_x y = (2x)(x^3 + 1) + (x^2 + 2)(3x^2) \]

Simplify the Expression

Compute each term:- First term: \((2x)(x^3 + 1) = 2x^4 + 2x\)- Second term: \((x^2 + 2)(3x^2) = 3x^4 + 6x^2\)Add these terms together:\[ D_x y = 2x^4 + 2x + 3x^4 + 6x^2 \]

Combine Like Terms

Combine the like terms in the expression:- Combine \(2x^4\) and \(3x^4\) to get \(5x^4\)- The expression becomes:\[ D_x y = 5x^4 + 6x^2 + 2x \]

Key concepts

Differentiation

Differentiation is a fundamental concept in calculus that deals with finding the rate at which a function is changing at any given point. It is used widely when analyzing curves and solving real-world problems related to rates of change.

The basic idea is to find the derivative of a function, which represents the slope of the tangent line to the curve of the function at any given point. For example, if we have a function that describes the position of a car over time, the derivative of that function would give us the car's instantaneous speed at any point in time.

Differentiation involves using specific rules and operations, such as the power rule, product rule, quotient rule, and chain rule, to calculate the derivatives. These rules help simplify the calculation process, especially when dealing with complex functions. Understanding differentiation allows you to model and predict changes effectively.

Function Derivatives

The derivative of a function provides crucial information about the function, including its rates of change and its behavior around specific points.

• The derivative of a function gives the slope of the function's graph at any point.
• It tells us how the function rises or falls, indicating turning points, maxima, and minima.
• This concept is essential for finding local extrema and optimizing and solving various engineering and physics problems.

For instance, if we have a quadratic function and find its derivative, we can determine where the function reaches its highest or lowest point. This is important for tasks such as finding the optimal dimensions for a certain shape or minimizing costs in production.

Differentiating functions can also help identify points of inflection where the concavity of the function changes. Knowing these characteristics allows us to create more precise models and make informed decisions.

Calculus Problem Solving

Solving calculus problems involves applying various differentiation techniques to uncover solutions for mathematical models and real-world applications. At the core, it requires identifying the correct method or rule that applies to the function at hand.

• The problem-solving process begins by clearly defining the given functions and choosing appropriate rules.
• The next step is to apply these rules accurately, such as the product or chain rules, and computing derivatives accordingly.
• The final part involves simplifying the results to obtain the derivative in its simplest form.

For more complex functions, breaking them down into simpler parts can be incredibly beneficial. By understanding each component, you can apply the correct rules to differentiate them efficiently. These steps help in translating the abstract concept of calculus into practical solutions that are understandable and applicable to real-world scenarios.

Successful calculus problem-solving requires practice and a clear understanding of its fundamental principles. With time, solving these problems enhances logical thinking and analytical skills, which are invaluable in academic and professional settings.