Problem 12 - Calculus for Biology and Medicine

Question

Differentiate the functions given with respect to the independent variable. $$ f(x)=2 x^{3} \cos \frac{\pi}{3}+\cos \frac{\pi}{6} $$

Step-by-Step Solution

Answer

The derivative of the function is \(3x^2\).

Step 1: Identify Constant Terms

We begin by identifying any constant terms in the function. The expressions \(\cos \frac{\pi}{3}\) and \(\cos \frac{\pi}{6}\) are constants as they do not depend on \(x\). Calculate their values: \(\cos \frac{\pi}{3} = \frac{1}{2}\) and \(\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}\). Thus, the function simplifies to \(f(x) = 2x^3 \cdot \frac{1}{2} + \frac{\sqrt{3}}{2}\).

Step 2: Simplify the Function

Simplify the given function using the constant values calculated in Step 1. The function simplifies to: \[ f(x) = x^3 + \frac{\sqrt{3}}{2} \].

Step 3: Differentiate the Simplified Function

Differentiate each term of the function with respect to \(x\). The derivative of \(x^3\) with respect to \(x\) is \(3x^2\). The derivative of \(\frac{\sqrt{3}}{2}\), a constant term, is \(0\).

Step 4: Write the Final Derivative

Combine the derivatives obtained in Step 3. The derivative of the function \(f(x) = x^3 + \frac{\sqrt{3}}{2}\) is \(f'(x) = 3x^2\).

Key Concepts

Derivatives

In calculus, the derivative of a function is a measure of how the function value changes as its input changes. The process of finding a derivative is called differentiation. To differentiate a function, we apply specific rules to each term. In the context of our exercise, where the function is \( f(x) = x^3 + \frac{\sqrt{3}}{2} \), we have two terms.

- The first term, \( x^3 \), changes as \( x \) changes; hence we take its derivative.- The second term, \( \frac{\sqrt{3}}{2} \), is a constant, and its derivative is zero.

The power rule, a fundamental rule of differentiation, tells us that to differentiate \( x^n \), we multiply by the power and reduce the power by one. This results in \( nx^{n-1} \). Applying this rule gives us the derivative of \( x^3 \) as \( 3x^2 \).

Thus, the derivative of our function simplifies to \( f'(x) = 3x^2 \), highlighting how the function's rate of change depends on \( x \).

Trigonometric Constants

Trigonometric constants are values obtained by evaluating trigonometric functions like sine and cosine at specific angles. In our exercise, we dealt with the expressions \( \cos \frac{\pi}{3} \) and \( \cos \frac{\pi}{6} \).These values arise frequently in trigonometry and are standard constants.

- \( \cos \frac{\pi}{3} = \frac{1}{2} \)- \( \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \)

Such constants simplify expressions by replacing trigonometric terms with numbers, making it easier to work with them. Understanding these constants is crucial because they help transform complex trigonometric expressions into simpler algebraic forms. This finds significant application when differentiating trigonometric functions as we convert them into simpler forms that are easier to handle for differentiation.

Simplifying Expressions

Simplifying mathematical expressions is a critical skill in calculus. It involves breaking down a complex function into simpler parts, which simplifies the process of differentiation. In the given exercise, we started with the function \( f(x)=2 x^{3} \cos \frac{\pi}{3} + \cos \frac{\pi}{6} \).

The first step is to identify constants in the function and substitute their respective values obtained from trigonometric identities. Converting \( \cos \frac{\pi}{3} \) to \( \frac{1}{2} \) and \( \cos \frac{\pi}{6} \) to \( \frac{\sqrt{3}}{2} \) helped us in simplifying the function to \( f(x) = x^3 + \frac{\sqrt{3}}{2} \).

By simplifying, the expression becomes much easier to differentiate.This process ensures that all terms of the function are in their simplest algebraic form, ready for straightforward application of differentiation rules.

Calculus Problem Solving

Solving calculus problems involves systematically applying mathematical rules and techniques to find solutions. Starting with functions that may seem complex, the key lies in decomposing them into more manageable parts.

In our scenario, we applied a sequence of logical steps:

• Identify trigonometric constants within the function: This step involved replacing the trigonometric parts with numerical values, using known constants.
• Simplify the function: Combining like terms by applying multiplication and addition.
• Apply differentiation: Use derivative formulas and rules to break down each part of the simplified function.
• Combine the results: Express the derivative in the simplest and most insightful form.

These steps ensure clarity and accuracy when solving calculus problems. By mastering these techniques, we can efficiently handle a wide variety of calculus-based tasks. The discipline lies in maintaining a clear and methodical approach, which is essential for success in calculus.