Problem 30 - Calculus for Biology and Medicine

Question

Differentiate the functions with respect to the independent variable. (Note that log denotes the logarithm to base 10.) $$ f(x)=\ln \left(1-x^{3}\right) $$

Step-by-Step Solution

Answer

\( \frac{d}{dx} f(x) = \frac{-3x^2}{1-x^3} \).

Step 1: Identify the Function

We are given the function \( f(x) = \ln(1 - x^3) \). We need to differentiate this function with respect to \( x \).

Step 2: Apply the Chain Rule

The function \( f(x) = \ln(1 - x^3) \) is a composition of functions where the outer function is \( g(u) = \ln(u) \) and the inner function is \( u(x) = 1 - x^3 \). To differentiate it, apply the chain rule: \( \frac{d}{dx} f(x) = \frac{d}{du} g(u) \cdot \frac{du}{dx} \).

Step 3: Differentiate Outer Function

Differentiate the outer function \( g(u) = \ln(u) \) with respect to \( u \): \( \frac{d}{du} \ln(u) = \frac{1}{u} \). So, \( \frac{d}{du} g(u) = \frac{1}{1-x^3} \) when substituting \( u = 1-x^3 \).

Step 4: Differentiate Inner Function

Differentiate the inner function \( u(x) = 1-x^3 \) with respect to \( x \): \( \frac{du}{dx} = -3x^2 \).

Step 5: Use the Chain Rule

Combine the results from Steps 3 and 4 using the chain rule: \( \frac{d}{dx} f(x) = \frac{1}{1-x^3} \cdot (-3x^2) = \frac{-3x^2}{1-x^3} \).

Step 6: Write the Final Result

The derivative of the function \( f(x) = \ln(1 - x^3) \) is \( \frac{d}{dx} f(x) = \frac{-3x^2}{1-x^3} \).

Key Concepts

Chain Rule

The Chain Rule is a vital tool in calculus used when differentiating composite functions. These are functions that are made up of one function inside another. In our exercise, we dealt with a composite function where the outer function is a logarithmic function and the inner is a polynomial function, specifically:

• Outer function, often labeled as \( g(u) \), is \( \ln(u) \).
• Inner function, referred to as \( u(x) \), is \( 1-x^3 \).

To apply the Chain Rule, differentiate the outer function with respect to its argument \( u \), then multiply by the derivative of the inner function with respect to \( x \).
Here's the process summarized:

• Find \( \frac{d}{du} \ln(u) = \frac{1}{u} \).
• Next, differentiate the inner function: \( \frac{du}{dx} = -3x^2 \).
• Finally, multiply these results: \( \frac{d}{dx} \ln(1-x^3) = \frac{1}{1-x^3} \cdot (-3x^2) \).

This leads us to the derivative \( \frac{-3x^2}{1-x^3} \). The Chain Rule is powerful, making it easier to handle complex derivatives that might appear daunting at first glance.

Logarithmic Differentiation

Logarithmic Differentiation involves using logarithms to simplify the differentiation of products, quotients, and powers in functions. It extensively utilizes the properties of logarithms to make differentiation more manageable. Logarithms convert multiplication into addition and division into subtraction, making them a perfect tool to tackle tricky differentiation problems.
In the exercise, logarithms helped in handling the complicated expression inside the logarithm function \( \ln(1-x^3) \). By differentiating the log function, we simplified the differentiation process.

• The derivative of \( \ln(u) \), where \( u = 1-x^3 \), is \( \frac{1}{u} \), symbolizing ease in dealing with division.
• Further utilizing, the chain rule gives us the complete derivative process.

Logarithmic Differentiation is incredibly effective in tackling functions with exponents or complex roots, providing clearer paths to derivative solutions that direct computation alone would not easily achieve.

Mathematical Functions

Mathematical functions are the backbone of calculus and different operations we perform in it, like differentiation. They define relationships between variables where each input (independent variable) has a unique output (dependent variable).
In our context, functions can be as simple as \( f(x) = x^2 \) or as complex as composite functions like \( f(x) = \ln(1 - x^3) \). Understanding the type of function we are dealing with is crucial for selecting the correct differentiation techniques:

• Linear functions: The simplest kind, involving no exponents or products, making straight differentiation possible.
• Polynomial functions: These involve sums of powers of the variable and require the power rule for differentiation.
• Exponential functions: Involving bases raised to a variable exponent, requiring special rules for differentiation.
• Logarithmic functions: Useful for simplifying products and powers before differentiation, as seen in our example.

The interplay of these functions and the correct application of differentiation rules allow us to navigate complex mathematical landscapes by finding derivatives and understanding rates of change within these functions.