Problem 21 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the indicated derivative. \(\frac{d z}{d x}\) 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 Function Components

The function given is \( z = x^2 \ln x^2 + (\ln x)^3 \). We can break it into two parts: \( z_1 = x^2 \ln x^2 \) and \( z_2 = (\ln x)^3 \).

Step 2: Find the Derivative of the First Term

For \( z_1 = x^2 \ln x^2 \), apply the product rule \( u'v + uv' \). Here, \( u = x^2 \) and \( v = \ln x^2 \). Compute \( u' = 2x \) and \( v' = \frac{d}{dx}(2 \ln x) = \frac{2}{x} \). Thus, the derivative is \( 2x \ln x^2 + x^2 \cdot \frac{2}{x} = 2x \ln x^2 + 2x \).

Step 3: Simplify the Derivative of the First Term

The term \( \ln x^2 \) can be rewritten using the logarithm property \( \ln x^2 = 2 \ln x \). Substitute this into the derivative to get \( 2x (2 \ln x) + 2x = 4x \ln x + 2x \).

Step 4: Find the Derivative of the Second Term

For \( z_2 = (\ln x)^3 \), use the chain rule. If \( u = \ln x \), then \( z_2 = u^3 \) and \( \frac{d}{dx}u^3 = 3u^2 \cdot u' \). Compute \( u' = \frac{1}{x} \). Thus, the derivative is \( 3(\ln x)^2 \cdot \frac{1}{x} = \frac{3(\ln x)^2}{x} \).

Step 5: Combine Derivatives of Both Terms

The derivative \( \frac{dz}{dx} = \) (derivative of first term) + (derivative of second term), which is \( 4x \ln x + 2x + \frac{3(\ln x)^2}{x} \).

Step 6: Simplify the Combined Derivative

Combine the terms to get \( \frac{dz}{dx} = 4x \ln x + 2x + \frac{3(\ln x)^2}{x} \) as a single expression.

Key Concepts

Product Rule

In calculus, the product rule is an essential technique for finding the derivative of a product of two functions. Suppose you have two functions, say \( u(x) \) and \( v(x) \). If you want to differentiate their product, \( u(x)v(x) \), the product rule tells you to do the following:

• First, differentiate \( u(x) \), resulting in \( u'(x) \).
• Then, differentiate \( v(x) \), resulting in \( v'(x) \).
• Combine them with the formula: \( u'(x)v(x) + u(x)v'(x) \).

For example, if \( z_1 = x^2 \ln x^2 \), you set \( u = x^2 \) and \( v = \ln x^2 \). The derivative becomes \( u'v + uv' \) which calculates to \( 2x \ln x^2 + 2x^2\cdot \frac{2}{x} \). This nifty rule is a powerful tool that allows you to differentiate complex multiplicative combinations efficiently.

Chain Rule

The chain rule is another fundamental method in calculus for finding derivatives, specifically when dealing with composite functions. Consider a function that is a composition of two functions, say \( h(x) = f(g(x)) \). To find \( h'(x) \), which is the derivative, follow these steps:

• Differentiate the outer function \( f \) with respect to the inner function \( g \), resulting in \( f'(g(x)) \).
• Multiply this by the derivative of the inner function \( g'(x) \).

In our exercise, to handle \( z_2 = (\ln x)^3 \), you recognize it as \( u^3 \) where \( u = \ln x \). Applying the chain rule, you find the derivative is \( 3u^2 \cdot u' \), where \( u' = \frac{1}{x} \). Therefore, the whole derivative becomes \( \frac{3(\ln x)^2}{x} \). This rule is a must-have in your calculus toolkit, particularly for nested functions.

Logarithmic Differentiation

Logarithmic differentiation is a clever method often used when differentiating complicated expressions, especially those containing powers or products of functions. The power of using logarithms is showcased through the simplification properties of logarithms.
Once you take a logarithm of both sides of an equation, the property \( \ln(a^b) = b\ln a \) allows you to rewrite and simplify, making differentiation much more straightforward.
In our specific example, the term \( \ln x^2 \) initially appears complex but can be rewritten using the identity \( \ln x^2 = 2 \ln x \). This simplification aids in straightforward differentiation and illustrates the power logarithms have in tackling otherwise cumbersome derivatives.

Simplifying Expressions

Simplifying expressions is a crucial step after finding derivatives, as it allows us to present the derivative in its most reduced and understandable form. Leaving the derivative in a complex state can lead to errors or confusion.
In our derivative \( \frac{dz}{dx} = 4x \ln x + 2x + \frac{3(\ln x)^2}{x} \), each term has been carefully combined. Initially, terms like \( 2x \ln x^2 \) were simplified using properties of logarithms and arithmetic to form \( 4x \ln x + 2x \). Simplified forms not only make mathematical expressions aesthetically pleasing but also easier to work with, especially in further calculations or evaluations.