Problem 1 - Calculus for Biology and Medicine

Question

Find the derivative with respect to the independent variable. $$ f(x)=2 \sin x-\cos x $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = 2 \cos x + \sin x \).

Step 1: Identify the Function

The given function is \( f(x) = 2 \sin x - \cos x \). We need to find its derivative with respect to \( x \).

Step 2: Recall the Derivative Rules

The derivative of \( \sin x \) with respect to \( x \) is \( \cos x \) and the derivative of \( \cos x \) with respect to \( x \) is \( -\sin x \).

Step 3: Differentiate Each Term

Apply the derivative rules to each term independently. For \( 2 \sin x \), by constant multiple rule, the derivative is \( 2 \cos x \). For \(- \cos x\), the derivative is \(-(-\sin x) = \sin x\).

Step 4: Write the Combined Derivative

Combine the results from Step 3 to write the derivative of the entire function: \( f'(x) = 2 \cos x + \sin x \).

Key Concepts

Trigonometric Functions

Trigonometric functions are foundational in calculus. They relate the angles and ratios of sides in right triangles to the unit circle. In calculus, we commonly work with the sine ( \( \sin \)) and cosine ( \( \cos \)) functions. These functions have periodical properties:

• \( \sin x \) represents the y-coordinate of a point on the unit circle, corresponding to an angle \( x \) from the positive x-axis.
• \( \cos x \) represents the x-coordinate of the same point on the unit circle.

These functions exhibit smooth, wave-like properties and we're often interested in their rates of change, which leads us to their derivatives. Understanding these underlying geometric interpretations helps when analyzing problems involving oscillations, waves, and periodic behavior in physics and engineering contexts.
Recognizing how to derive trigonometric functions is a key skill that extends beyond simple graph interpretations to solving more complex calculus problems.

Differentiation Rules

In calculus, differentiation is the process of finding the derivative of a function. This represents the rate at which the function's value changes with respect to changes in its input. To solve such problems efficiently, understanding differentiation rules is crucial:

• Basic Trigonometric Derivatives: The derivative of \( \sin x \) is \( \cos x \), and the derivative of \( \cos x \) is \( -\sin x \).
• Constant Multiple Rule: If you have a function multiplied by a constant, like \( c \cdot f(x) \), its derivative is simply \( c \cdot f'(x) \). For example, \( 2\sin x \) differentiates to \( 2 \cdot \cos x \).
• Sum Rule: The derivative of a sum of functions, such as \( f(x) + g(x) \), is the sum of their derivatives: \( f'(x) + g'(x) \).

Armed with these rules, differentiation becomes a systematic process. Instead of evaluating each term in a lengthy fashion, recognizing and applying the rules efficiently allows students to arrive at solutions quickly.

Calculus Problem Solving

Solving calculus problems involves a step-by-step approach, analyzing, applying rules, and verifying results at each stage. Here are some tips to enhance your calculus problem-solving experience:

• Understand the Problem: Start by identifying the function and what the problem is asking for. In our case, we had to differentiate \( f(x) = 2 \sin x - \cos x \).
• Apply Differentiation Rules: Use the differentiation rules deliberately. Breaking down the function into parts, differentiate each term according to the applicable rules, and ensure constants are handled with the constant multiple rule.
• Simplify Where Possible: Once you've found the derivative, simplify the result when possible. This final expression, \( f'(x) = 2 \cos x + \sin x \), should reflect the simplest form of the derivative.
• Verify with the Original Function: Look back at the original function to ensure all aspects have been accounted for, and reconfirm that all rules were applied correctly.

By following these steps, students can tackle calculus problems with confidence. Starting with a clear structure provides a solid foundation for more complex problem-solving scenarios in calculus.