Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the indicated derivative. \(\frac{d y}{d x}\) if \(y=x^{2} \ln x\)

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = 2x \ln x + x \).

Step 1 - Identify the Rule

To find the derivative of the function \( y = x^2 \ln x \), we recognize that this is a product of two functions: \( u = x^2 \) and \( v = \ln x \). This requires using the product rule for differentiation, which states that if \( y = u \, v \), then \( \frac{dy}{dx} = u'v + uv' \).

Step 2 - Differentiate First Function (\(u\))

Differentiate \( u = x^2 \) with respect to \( x \). The derivative of \( x^n \) is given by \( nx^{n-1} \). So, \( u' = 2x \).

Step 3 - Differentiate Second Function (\(v\))

Differentiate \( v = \ln x \) with respect to \( x \). The derivative of \( \ln x \) is \( \frac{1}{x} \). Therefore, \( v' = \frac{1}{x} \).

Step 4 - Apply Product Rule

Substitute \( u' \), \( v \), \( u \), and \( v' \) into the product rule formula: \[ \frac{dy}{dx} = u'v + uv' = (2x)(\ln x) + (x^2)\left(\frac{1}{x}\right).\]

Step 5 - Simplify the Expression

Simplify the expression from Step 4: \[ \frac{dy}{dx} = 2x \ln x + x.\] This is the derivative of \( y = x^2 \ln x \).

Key Concepts

Derivative Calculation

The process of finding the derivative of a function is fundamental in calculus. Differentiation allows us to determine the rate at which a function changes at any given point. Calculating derivatives involves rules and techniques that can be applied to various types of functions, such as polynomials, trigonometric functions, and logarithmic functions.

When dealing with products of functions, like in the provided exercise, we often use the product rule of differentiation. This rule is particularly useful when working with expressions that combine different types of functions. The product rule states that if you have a function composed of two smaller functions, say \( y = u \cdot v \), then the derivative of \( y \) with respect to \( x \) is calculated as

• \( \frac{dy}{dx} = u'v + uv' \)

where \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \) respectively. This rule simplifies the process of differentiation in situations where functions are multiplied, such as in the exercise with the function \( y = x^2 \ln x \). By breaking down the function into parts, differentiating each part, and then combining them, you can effectively find the derivative.

Natural Logarithm Differentiation

Natural logarithms (\( \ln x \)) have their unique rules when it comes to differentiation. The derivative of the natural logarithm function is relatively straightforward, but it requires special attention. For any function \( \ln x \), the derivative is given by

• \( \frac{d}{dx}(\ln x) = \frac{1}{x} \).

This means that when calculating the derivative of any expression involving \( \ln x \), one must apply this specific rule.

In the given exercise, the function \( v = \ln x \) is a natural logarithm. We differentiate \( v \) by applying the rule, resulting in \( v' = \frac{1}{x} \). This result is then used along with the product rule to find the overall derivative of the function \( y = x^2 \ln x \). Understanding how to differentiate natural logarithms is crucial for correctly applying the product rule and obtaining the right derivative.

Polynomial Function Differentiation

Differentiating polynomial functions is one of the basic skills to master in calculus. Polynomials are expressions that include terms like \( x^n \), where \( n \) is a non-negative integer. The rule for differentiating a polynomial term \( x^n \) is given by:

• \( \frac{d}{dx}(x^n) = nx^{n-1} \)

This rule implies that to find the derivative of each term, you multiply by the current power and then subtract one from the exponent.

In the original problem, we were asked to differentiate \( u = x^2 \), a polynomial function, with respect to \( x \). According to the polynomial differentiation rule, the derivative of \( x^2 \) is \( 2x \). This result is crucial when applying the product rule for finding the derivative of the entire function \( y = x^2 \ln x \). Having a firm grasp on polynomial differentiation equips you with the tools to handle more complex calculus problems involving products of mixed-function types.