Problem 28 - Calculus for Biology and Medicine

Question

Differentiate the functions with respect to the independent variable. (Note that log denotes the logarithm to base 10.) $$ f(x)=\ln \left(1-x^{2}\right) $$

Step-by-Step Solution

Answer

The derivative of \( f(x)=\ln(1-x^2) \) is \( f'(x) = \frac{-2x}{1-x^2} \).

Step 1: Identify the Composition of Functions

The function given is composed of a natural logarithm function and a polynomial function. Specifically, we have the natural log function applied to \( g(x) = 1-x^2 \). Therefore, we will need to use the chain rule to differentiate \( f(x) = \ln(g(x)) \).

Step 2: Apply the Chain Rule

The chain rule states that if a function \( y = \ln(u) \), then its derivative \( \frac{dy}{dx} \) is given by \( \frac{1}{u} \frac{du}{dx} \). Therefore, \( f'(x) = \frac{1}{1-x^2} \cdot \frac{d}{dx}(1-x^2) \).

Step 3: Differentiate the Inner Function

Differentiate the inner function \( g(x) = 1-x^2 \). The derivative is \( \frac{d}{dx}(1-x^2) = -2x \).

Step 4: Combine Using the Chain Rule

Substitute the derivative of the inner function back into the expression from Step 2. Thus, we have \( f'(x) = \frac{1}{1-x^2} \cdot (-2x) \), which simplifies to \( f'(x) = \frac{-2x}{1-x^2} \).

Key Concepts

Understanding the Chain Rule

The chain rule is a fundamental technique in calculus used to differentiate functions that are composed of other functions. It simplifies the process of differentiation when dealing with complex functions by breaking them down into simpler components. Here's how the chain rule works: If you have a composite function, say \( f(g(x)) \), where \( f \) is a function of \( g(x) \) and \( g(x) \) is a function of \( x \), the derivative of this composite function is found by:

• First, differentiating the outer function, \( f \), with respect to its inner function, \( u = g(x) \).
• Then, taking the derivative of the inner function, \( g(x) \), with respect to \( x \).
• Finally, multiplying these two derivatives together.

In mathematical terms, it's written as: \[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \, g'(x) \] This rule is powerful for handling differentiations involving nested or composite functions, much like the example given with the natural logarithm function applied to a polynomial in the exercise.

Natural Logarithm Function

The natural logarithm, denoted as \( \ln(x) \), is a special logarithm that has the base \( e \), where \( e \) is approximately 2.71828. It's a key concept in many areas of mathematics, especially calculus, due to its unique properties and relationship with exponential functions. When differentiating a natural logarithm, it's important to remember these essential points:

• The derivative of \( \ln(u) \), where \( u \) is a differentiable function of \( x \), is \( \frac{1}{u} \frac{du}{dx} \). This highlights the need to use the chain rule when \( u \) is not simply \( x \), but another function.
• For a simple \( \ln(x) \), the derivative is \( \frac{1}{x} \).

Natural logarithms are particularly useful in integrating functions and solving differential equations due to their straightforward properties, making them a staple in algebra and calculus.

Polynomial Functions

Polynomial functions are expressions composed of variables raised to whole-number exponents and multiplied by coefficients. These functions are fundamental in mathematics and appear in many different forms, from simple linear functions like \( f(x) = x \), to more complex forms like \( f(x) = x^2 + 3x + 2 \).Differentiating polynomial functions is a straightforward process, often taught early in calculus:

• Each term in the polynomial is differentiated individually.
• The rule for differentiating \( x^n \) is to multiply by the current power of \( x \), \( n \), and then reduce that power by one, resulting in \( nx^{n-1} \).

Using this approach, a polynomial like \( g(x) = 1 - x^2 \) is simple to differentiate:- The derivative of the constant, 1, is 0.- For the term \( -x^2 \), the derivative is \( -2x \).This makes polynomial functions a great starting point for learning calculus concepts, as they introduce key differentiation techniques that are essential for more complex functions.