Problem 48 - Calculus for Biology and Medicine

Question

Differentiate the functions with respect to the independent variable. $$ f(x)=x^{2} \ln x^{2} $$

Step-by-Step Solution

Answer

The derivative of \(f(x) = x^2 \ln x^2\) is \(4x \ln x + 2x\).

Step 1: Identify the Problem

We need to find the derivative of the function \(f(x) = x^2 \ln x^2\) with respect to \(x\). This involves applying differentiation rules to a product of functions.

Step 2: Apply Product Rule

The product rule for differentiation states that the derivative of \(u(x)v(x)\) is \(u'(x)v(x) + u(x)v'(x)\). Identify \(u(x) = x^2\) and \(v(x) = \ln x^2\). We need to differentiate both \(u(x)\) and \(v(x)\) individually.

Step 3: Differentiate First Function

Differentiate \(u(x) = x^2\) with respect to \(x\). Using the power rule, \(u'(x) = 2x\).

Step 4: Differentiate Second Function

For \(v(x) = \ln x^2\), use the chain rule. Start by rewriting it as \(v(x) = 2 \ln x\). The derivative of \(\ln x\) is \(\frac{1}{x}\), so \(v'(x) = 2 \cdot \frac{1}{x} = \frac{2}{x}\).

Step 5: Combine Results Using Product Rule

Substitute \(u(x)\), \(u'(x)\), \(v(x)\), and \(v'(x)\) into the product rule formula: \(u'(x)v(x) + u(x)v'(x) = (2x)(\ln x^2) + (x^2)(\frac{2}{x})\). Simplify this to \(2x \ln x^2 + 2x\).

Step 6: Simplify the Expression

Simplifying further, recognize that \(\ln x^2 = 2 \ln x\). The derivative becomes \(4x \ln x + 2x\). This is the final derivative.

Key Concepts

Product Rule

When dealing with functions that are products of two (or more) functions, the product rule helps us find their derivative more efficiently. The product rule is formulated as follows:

• If you have two functions, say \( u(x) \) and \( v(x) \), their derivative as a product is given by \( (uv)' = u'v + uv' \).
• Essentially, you take the derivative of the first function, multiply it by the second function, then do the reverse, adding them together.

In our problem, we used the product rule to differentiate \(f(x) = x^2 \ln x^2\). We identified \(u(x) = x^2\) and \(v(x) = \ln x^2\). The task was to differentiate each individually before recombining them using the product rule formula. Through this method, we structure our work clearly and avoid confusion.

Chain Rule

The chain rule is a powerful tool in differentiation, especially when differentiating compositions of functions. It is used to differentiate composite functions, which are essentially functions inside of other functions. Its formula is:

• Suppose we have a composite function \( g(f(x)) \). Then its derivative is given by \( g'(f(x)) \, f'(x) \).
• It allows us to "chain" together the derivatives of the outer and inner functions.

In the solution, the chain rule was applied to break down the differentiation of \(v(x) = \ln x^2\) into simpler steps by expressing it as \(2 \ln x\). By understanding that \( x^2 \) inside the logarithm could be expressed more simply, we were able to find the derivative of \(\ln x = \frac{1}{x}\), then multiply by 2, according to the new formulation. This illustrates the chain rule perfectly, turning a complex derivative into a manageable one.

Power Rule

The power rule is perhaps the simplest and most commonly used rule in calculus when it comes to differentiation. It's used to find the derivative of powers of variables, and its formula is:

• For any function of the form \( f(x) = x^n \), the derivative is \( f'(x) = nx^{n-1} \).
• It's direct and intuitive, allowing quick calculations.

In the exercise, the power rule was specifically applied to differentiate \(u(x) = x^2\). According to the power rule, the derivative is simply \( 2x \) because the original function was \( x^2 \). This straightforward application demonstrates why the power rule is favored for polynomial expressions: it's quick and effective, simplifying complicated derivatives into simple arithmetic.