Problem 21 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \underline{\phantom{xxx}} , \text { find the indicated derivative. } $$ $$ \frac{d z}{d x} \text { if } z=x^{2} \ln x^{2}+(\ln x)^{3} $$

Step-by-Step Solution

Answer

\( \frac{dz}{dx} = 4x \ln x + 2x + \frac{3(\ln x)^2}{x} \).

Step 1: Identify the Derivative Components

The function to differentiate is given by \( z = x^2 \ln(x^2) + (\ln x)^3 \). The expression consists of two parts: \( x^2 \ln(x^2) \) and \((\ln x)^3\). We need to find the derivative of each part separately.

Step 2: Differentiate the First Term

The first term is \( x^2 \ln(x^2) \). Applying the product rule \( (uv)' = u'v + uv' \), where \( u = x^2 \) and \( v = \ln(x^2) \):- \( u' = 2x \)- \( v = \ln(x^2) = 2\ln(x) \), so \( v' = \frac{2}{x} \)Thus, the derivative of the first term is \( u'v + uv' = (2x)(2\ln x) + (x^2)(\frac{2}{x}) = 4x\ln x + 2x \).

Step 3: Differentiate the Second Term

The second term is \( (\ln x)^3 \). Use the Chain Rule \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \), where \( f(u) = u^3 \) and \( g(x) = \ln x \):- \( f'(u) = 3u^2 \)- \( g'(x) = \frac{1}{x} \)Apply the chain rule: \( 3(\ln x)^2 \cdot \frac{1}{x} = \frac{3(\ln x)^2}{x} \).

Step 4: Combine the Derivatives

Combine the derivatives of the two parts to find \( \frac{dz}{dx} \):\[ \frac{dz}{dx} = 4x\ln x + 2x + \frac{3(\ln x)^2}{x} \].

Step 5: Simplify the Expression

Simplify the expression to get the final derivative:\[ \frac{dz}{dx} = 4x \ln x + 2x + \frac{3(\ln x)^2}{x} \]. Combine terms if possible, but the expression is already in simplified form.

Key Concepts

Product Rule

The product rule is a fundamental tool in calculus that is used when differentiating expressions that are the product of two functions. For two functions, \( u(x) \) and \( v(x) \), the rule states that the derivative of their product is given by:\[(uv)' = u'v + uv'\]This means that you take the derivative of the first function \( u \) and multiply it by the second function \( v \), then add the product of the first function and the derivative of the second function. For instance, consider the term \( x^2 \ln(x^2) \) in our expression. Here, \( u = x^2 \), which differentiates to \( u' = 2x \), and \( v = \ln(x^2) \), which simplifies to \( 2\ln(x) \) and differentiates to \( v' = \frac{2}{x} \). Applying the product rule helps us systematically find the derivative in steps.

• Multiply \( u' \) with \( v \)
• Multiply \( u \) with \( v' \)
• Obtain the derivative as \( 4x\ln(x) + 2x \)

Mastering the product rule makes handling complex expressions like this much more manageable.

Chain Rule

The chain rule is another vital rule that helps us differentiate composite functions. In simple terms, if you have a function nested within another function, the chain rule comes into play. For the composite function \( f(g(x)) \), the derivative can be expressed as:\[(f(g(x)))' = f'(g(x)) \cdot g'(x)\]This rule tells us to first take the derivative of the outer function while leaving the inner function unchanged, then multiply it by the derivative of the inner function.
A typical example from our exercise is the term \((\ln x)^3\). Here:

• Consider \( f(u) = u^3 \), with \( u = \ln x \), so \( f'(u) = 3u^2 \).
• Also, differentiate \( g(x) = \ln x \) to find \( g'(x) = \frac{1}{x} \).
• Apply the chain rule to get \( \frac{3(\ln x)^2}{x} \).

With practice, the chain rule becomes a powerful method to effortlessly differentiate complex functions.

Logarithmic Differentiation

Logarithmic differentiation is a unique technique especially helpful when dealing with products, quotients, or powers. It utilizes properties of logarithms to simplify the differentiation process, making it particularly useful when trying to avoid cumbersome calculations.
To use this technique:

• Take the natural logarithm (\( \ln \)) of both sides.
• Differentiating using the properties of logarithms simplifies the process.
• Apply other differentiation rules such as the product or chain rule where appropriate.

Although logarithmic differentiation isn't directly illustrated in our current problem, it becomes critical in handling more complex expressions, providing a way to reduce heavy algebraic manipulations, especially with exponential and logarithmic terms present in our functions, as we've seen in simplifying \( \ln(x^2) \) to ease the differentiation. Utilizing this method can save a lot of time and effort during calculation.