Problem 53 - Calculus for Biology and Medicine

Question

Differentiate the functions with respect to the independent variable. (Note that log denotes the logarithm to base 10.) $$ f(x)=\ln \left|x^{2}-3\right| $$

Step-by-Step Solution

Answer

The derivative is \( \frac{2x}{x^2 - 3} \).

Step 1: Understanding the Function

The function provided is a logarithmic function using natural logarithm: \( f(x) = \ln |x^2 - 3| \). Our goal is to find the derivative of this function with respect to \( x \).

Step 2: Recognize the Chain Rule Requirement

The function inside the natural logarithm, \( |x^2 - 3| \), requires us to use the chain rule in calculus, which involves differentiating the outer function (\( \ln u \)) with respect to the inner function \( u = |x^2 - 3| \), and then differentiating \( u \) with respect to \( x \).

Step 3: Differentiate the Outer Function

The derivative of \( \ln u \) with respect to \( u \) is \( \frac{1}{u} \). We apply this to the outer function, giving a partial derivative of \( \frac{1}{|x^2 - 3|} \).

Step 4: Differentiate the Inner Function using Absolute Value Rule

We differentiate \( |x^2 - 3| \) with respect to \( x \). The derivative of \( x^2 - 3 \) is \( 2x \). Since we have an absolute value, we incorporate the sign of the function as \( \frac{x}{|x|} \), giving us \( 2x \cdot \frac{x^2 - 3}{|x^2 - 3|} \).

Step 5: Apply the Chain Rule

Now, by applying the chain rule: \( \left( \frac{df}{dx} = \frac{1}{|x^2 - 3|} \cdot 2x \cdot \frac{x^2 - 3}{|x^2 - 3|} \right) \). This simplifies to \( \frac{2x}{x^2 - 3} \).

Step 6: Simplify the Final Expression

As the absolute value and the function cancel out due to the expression \( 2x \cdot \frac{x^2 - 3}{|x^2 - 3|} \) simplifying, the final derivative simplifies to \( \frac{2x}{x^2 - 3} \).

Key Concepts

Chain Rule

In calculus, the chain rule is a fundamental technique used for differentiating composite functions. This rule becomes essential when dealing with functions that have an "inner function" and an "outer function". For example, if you have a function like \( f(g(x)) \), where \( f \) is your outer function and \( g(x) \) is your inner function, the chain rule helps us differentiate it correctly.

Using the chain rule involves two steps:

• First, differentiate the outer function \( f \) with respect to the inner function \( g(x) \).
• Then, multiply what you've found by the derivative of the inner function \( g(x) \) with respect to \( x \).

This is why it's called the "chain" rule because it's essentially linking the derivatives together. Applying this to the given problem, we see how the chain rule is used to differentiate \( \ln |x^2 - 3| \), where \( \ln u \) is the outer function and \( |x^2 - 3| \) is the inner function.

Natural Logarithm

Natural logarithms, denoted as \( \ln \), are logarithms with base \( e \), where \( e \approx 2.718 \.\.\. \). This constant \( e \) is an irrational number very important in mathematics, particularly in calculus and exponential functions.

When it comes to differentiation, the derivative of the natural logarithm of a function \( \ln(u) \) is quite straightforward. It's given by the rule:

• \( \frac{d}{dx} \ln(u) = \frac{1}{u} \cdot \frac{du}{dx} \)

This means you first take the reciprocal of the inner function \( u \) and then multiply by the derivative of \( u \). For our specific task, \( u = |x^2 - 3| \), so initially, we find \( \frac{1}{|x^2 - 3|} \) and later multiply it with the derivative of \( |x^2 - 3| \). It’s fascinating how natural logs can simplify calculus problems!

Absolute Value Differentiation

Differentiating functions that include absolute values can be tricky, but there's a solution. The absolute value of a function \( |v(x)| \) affects differentiation because the graph of \( |v(x)| \) changes its slope depending on whether the values are positive or negative.

When you're faced with differentiating \( |x^2 - 3| \), divide the process:

• Determine the derivative of the function \( v(x) = x^2 - 3 \), which is \( 2x \).
• Then, consider the effect of the absolute value, which can be expressed through the formula \( \frac{d}{dx} |v(x)| = v'(x) \cdot \frac{v(x)}{|v(x)|} \).

This means you adjust for the sign of \( x^2 - 3 \) by multiplying by this factor. Consequently, the derivative of \( |x^2 - 3| \) is \( 2x \cdot \frac{x^2 - 3}{|x^2 - 3|} \). In the end, the absolute value helps to refine where the function can change direction, crucial for fully simplifying the derivative.