Problem 5 - Calculus for Biology and Medicine

Question

Find the derivative with respect to the independent variable. $$ f(x)=\cos (x+1) $$

Step-by-Step Solution

Answer

The derivative is \(-\sin(x+1)\).

Step 1: Identify the formula for derivative

The function given is \( f(x) = \cos(x+1) \). To find its derivative, we apply the chain rule: \( \frac{d}{dx}[\cos(u)] = -\sin(u) \cdot \frac{du}{dx} \). Here, \( u = x + 1 \).

Step 2: Find the derivative of the inner function

Since \( u = x + 1 \), the derivative of \( u \) with respect to \( x \) is \( \frac{du}{dx} = 1 \).

Step 3: Apply the chain rule

Using the chain rule from Step 1, the derivative of \( f(x) \) becomes \( f'(x) = -\sin(x+1) \cdot \frac{du}{dx} \).

Step 4: Substitute the inner function derivative

Substitute the derivative of the inner function from Step 2 into the equation from Step 3: \( f'(x) = -\sin(x+1) \cdot 1 = -\sin(x+1) \).

Key Concepts

Understanding the Chain Rule

The chain rule is a fundamental concept in calculus used to find the derivative of composite functions. A composite function is when one function is applied to the result of another function. In simpler terms, it's a "function within a function." For example, the function \( f(x) = \cos(x + 1) \) is composite because it has an inner function, \( u = x + 1 \), inside the outer cosine function.
When using the chain rule, our goal is to differentiate the outer function with respect to the inner function and then multiply by the derivative of the inner function itself. In symbolic form, if you have a function \( y = g(f(x)) \), its derivative is given by:

• \( \frac{dy}{dx} = \frac{dg}{df} \cdot \frac{df}{dx} \)

In our example, \( f(u) = \cos(u) \) and \( u = x + 1 \). Thus, by applying the chain rule:

• First, differentiate the outer function \( \cos(u) \) to get \( -\sin(u) \).
• Then, find the derivative of the inner function \( u = x + 1 \), which is simply \( 1 \).
• Multiplying these together gives us \(-\sin(x + 1) \cdot 1\), resulting in \(-\sin(x + 1)\).

This is how the chain rule makes it manageable to tackle the derivatives of more complex functions.

Trigonometric Functions and Their Derivatives

Trigonometric functions include sine, cosine, and tangent functions, and they are foundational in mathematics, especially in calculus. Knowing how to differentiate these functions is crucial. Let's focus on the differentiation aspect of these functions, particularly for cosine:

• The derivative of \( \cos(u) \) with respect to \( u \) is \( -\sin(u) \).

This negative sign is important, as trigonometric function derivatives often differ slightly, making them sometimes tricky to remember. By learning these differentiations, one can handle a variety of problems involving periodic or wave-like functions.
In the context of our problem, we used this differentiation to help find the derivative of \( f(x) = \cos(x+1) \). By identifying the inner part \( x+1 \), differentiating \( \cos(u) \) as \( -\sin(u) \), and applying the chain rule, we determine that the derivative becomes \( -\sin(x+1) \). Understanding the behavior of trigonometric functions and their derivatives is essential for solving numerous calculus problems, from simple to complex.

The Basics of Calculus and Its Application

Calculus is a branch of mathematics focused on the concepts of change and motion. The two fundamental aspects of calculus are differentiation and integration, with differentiation primarily concerned with rates of change and the slopes of curves. In this exercise, we concentrated on differentiation, aiming to find the derivative of a specific function. Differentiation helps answer questions like, "How fast is something changing at a given point?"
The derivative of a function, in simple terms, provides us with a way to measure the function's rate of change. For the function \( f(x) = \cos(x + 1) \), we used the chain rule to find the derivative: \( f'(x) = -\sin(x + 1) \). This derivative tells us the rate at which the function's value is changing with respect to \( x \).
Why is this important? Derivatives have countless applications:

• They help in finding maxima and minima of functions, essential for optimization problems.
• They are used in physics to understand motion, providing insights into velocity and acceleration.
• They are fundamental in economics to determine profit maximization and cost minimization.

In essence, calculus allows us to understand and model real-world phenomena, making it a vital tool in mathematics and science.