Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \text { In Problems } 1-44, \text { find } D_{x} y \text { using the rules of this section. } $$ $$ y=x\left(x^{2}+1\right) $$

Step-by-Step Solution

Answer

The derivative is \( 3x^2 + 1 \).

Step 1: Identify the function components

The given function is \( y = x(x^2 + 1) \). It can be thought of as a product of two functions: \( u(x) = x \) and \( v(x) = x^2 + 1 \).

Step 2: Apply the Product Rule

The derivative of a product \( u(x)v(x) \) is given by \( D_x[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \). Identify \( u(x) = x \) and \( v(x) = x^2 + 1 \).

Step 3: Find the derivatives of u(x) and v(x)

Calculate \( u'(x) \) and \( v'(x) \).\( u(x) = x \) implies \( u'(x) = 1 \).\( v(x) = x^2 + 1 \) implies \( v'(x) = 2x \).

Step 4: Substitute and simplify

Substitute \( u'(x), v(x), u(x), \) and \( v'(x) \) into the product rule:\[ D_x [x(x^2 + 1)] = 1 \cdot (x^2 + 1) + x \cdot (2x) \]Simplify the expression:\[ = x^2 + 1 + 2x^2 = 3x^2 + 1 \].

Step 5: Write the final result

The derivative of \( y \) with respect to \( x \) is \( D_x(y) = 3x^2 + 1 \).

Key Concepts

Product Rule

The product rule is a crucial technique in calculus used for differentiating products of functions. Imagine you want to find the derivative of two multiplied functions, such as \( y = x(x^2 + 1) \). To do this, you need the product rule principle.

When you have two functions multiplied together, say, \( u(x) \) and \( v(x) \), the derivative is given by:

• \( D_x[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \)

This means you find the derivative of each function, \( u(x) \) and \( v(x) \), separately. Then, you substitute them into the formula. It's like filling in the blanks: take the derivative of the first function times the second function, plus the first function times the derivative of the second.

This handy tool works for any two functions being multiplied, helping you break down complex problems into more manageable parts.

Differentiation

Differentiation is the process of finding a derivative, which is a measure of how a function changes as its input changes. It tells us the rate of change, or slope, at any given point on a graph.

For the function \( y = x(x^2 + 1) \), differentiation helps determine how \( y \) changes with respect to \( x \). To do this effectively, especially when dealing with products of functions, you should use rules like the product rule. Differentiation is fundamental because:

• It allows us to calculate instantaneous rates of change, like velocity or growth rates.
• It's used to find the slope of a tangent line to a curve at any point.
• In calculus, it is essential for finding optimum points, like maxima and minima.

Each type of function has its own rules for differentiation. For instance, \( x^n \) has a simple rule: the derivative is \( nx^{n-1} \).

Using these rules helps you navigate through various problems, making differentiation a powerful tool for mathematicians and scientists alike.

Functions

Functions are mathematical entities that relate an input to an output. In calculus, functions play a crucial role since they describe relationships between variables.

For example, consider the function \( y = x(x^2 + 1) \). Here, \( y \) is dependent on \( x \), making \( x \) the independent variable. Functions like these can represent real-life phenomena, like the trajectory of a ball or the growth of a plant.

Understanding functions involves:

• Identifying how the output changes as the input changes.
• Recognizing different types such: linear, quadratic, polynomial, etc.
• Decomposing complex functions into simpler ones to aid in problem-solving. For instance, breaking \( y = x(x^2 + 1) \) into \( u(x) = x \) and \( v(x) = x^2 + 1 \).

Functions are vital because they provide a way to model and solve real-world problems effectively. Knowing how to manipulate and work with them opens the door to deeper mathematical analysis and understanding.