Problem 48 - Calculus for Biology and Medicine

Question

Differentiate the functions with respect to the independent variable. (Note that log denotes the logarithm to base 10.) $$ f(x)=x^{2} \ln x^{2} $$

Step-by-Step Solution

Answer

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

Step 1: Identify Parts of the Function

The function is given by \( f(x) = x^2 \ln(x^2) \). We need to differentiate this function with respect to \( x \). This function is a product of two functions: \( u(x) = x^2 \) and \( v(x) = \ln(x^2) \).

Step 2: Differentiate Using the Product Rule

Using the product rule for differentiation, which states that \((uv)' = u'v + uv'\), we need to differentiate \( u(x) = x^2 \) and \( v(x) = \ln(x^2) \) separately.

Step 3: Differentiate \( u(x) = x^2 \)

Differentiate the function \( u(x) = x^2 \) using the power rule, yielding \( u'(x) = 2x \).

Step 4: Differentiate \( v(x) = \ln(x^2) \)

To differentiate \( v(x) = \ln(x^2) \), use the chain rule. The derivative of \( \ln(u) \) with respect to \( u \) is \( 1/u \), so \( \ln(x^2) \) becomes \( v'(x) = \frac{1}{x^2} \cdot 2x = \frac{2}{x} \).

Step 5: Apply the Product Rule

Substitute \( u'(x) \), \( v(x) \), \( u(x) \), and \( v'(x) \) into the product rule formula: \( (uv)' = u'v + uv' \). This gives \( f'(x) = (2x) \ln(x^2) + x^2 \left(\frac{2}{x}\right) \).

Step 6: Simplify the Expression

Simplify the expression: \( f'(x) = 2x \ln(x^2) + 2x \). This simplifies the derivative fully.

Key Concepts

Product Rule

The product rule is a handy tool in calculus used to differentiate products of two or more functions. When you are given a function that is a product of two functions, say \( u(x) \) and \( v(x) \), the derivative is not as simple as taking the derivative of each individually. Instead, the product rule states that the derivative of \( u(x)v(x) \) is:

• \((uv)' = u'v + uv'\)

This means you must first differentiate each function individually to find \( u'(x) \) and \( v'(x) \), and then apply the formula.
In our example, the function \( f(x) = x^2 \ln(x^2) \) is the product of \( u(x) = x^2 \) and \( v(x) = \ln(x^2) \). By finding \( u'(x) = 2x \) and \( v'(x) = \frac{2}{x} \), we apply the product rule:

• \( (2x) \ln(x^2) + x^2 \left(\frac{2}{x}\right) \)

After simplifying, this results in the derivative \( f'(x) = 2x \ln(x^2) + 2x \).

Power Rule

The power rule is one of the simplest and most frequently used differentiation rules in calculus. It is specifically used for functions of the form \( x^n \), where \( n \) is any real number. The power rule states:

• The derivative of \( x^n \) is \( nx^{n-1} \)

This rule makes differentiating polynomial functions much easier. In our given function, \( u(x) = x^2 \), you can quickly apply the power rule and find the derivative:

• \( u'(x) = 2x^{2-1} = 2x \)

Recognizing when to apply the power rule simplifies many differentiation problems, saving time and effort. Always remember, apply the power rule directly to find quick derivative solutions for polynomial terms.

Chain Rule

The chain rule is vital when differentiating complex functions that can be viewed as a composition of two or more functions. Whenever you have a function wrapped inside another function, like \( g(h(x)) \), differentiation requires the chain rule. The chain rule formula is:

• If \( y = g(h(x)) \), then \( \frac{dy}{dx} = g'(h(x)) \cdot h'(x) \)

In our exercise, \( v(x) = \ln(x^2) \) can be seen as the composition of \( \ln(u) \) where \( u = x^2 \). To differentiate \( \ln(x^2) \), first differentiate the outer \( \ln(u) \) yielding \( \frac{1}{u} \), and then differentiate \( u = x^2 \) to get \( 2x \). Thus using the chain rule, we derive:

• \( v'(x) = \frac{1}{x^2} \cdot 2x = \frac{2}{x} \)

This method of breaking down functions into simpler parts via the chain rule is crucial for a clear and straightforward differentiation process.