Problem 6 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify Components for Chain Rule

The function given is \( f(x) = \sin(2-x) \). To differentiate this function, we'll use the chain rule, which applies when differentiating composite functions. Here, the outer function is \( \, \sin(u) \) and the inner function is \( \, u = 2-x \).

Step 2: Differentiate the Outer Function

Differentiate the outer function \( \, \sin(u) \) with respect to \( u \). The derivative of \( \, \sin(u) \) is \( \, \cos(u) \).

Step 3: Differentiate the Inner Function

Next, differentiate the inner function \( \, u = 2-x \) with respect to \( x \). The derivative of \( \, 2-x \) is \( \, -1 \).

Step 4: Apply the Chain Rule

Using the chain rule, multiply the derivative of the outer function \( \, \cos(u) \) by the derivative of the inner function \( \, -1 \). This gives us \( \, \cos(2-x) \times (-1) \).

Step 5: Simplify the Expression

Simplify the expression by performing the multiplication, resulting in \( \, -\cos(2-x) \). Thus, the derivative of \( f(x) = \sin(2-x) \) is \( f'(x) = -\cos(2-x) \).

Key Concepts

Chain Rule

The chain rule is an essential tool in calculus used to differentiate composite functions. Think of it as a method that lets us find the derivative of a function that contains another function within it. To apply the chain rule, you begin by identifying the two key components: an outer function and an inner function.
For instance, if you have a function like \( f(x) = \sin(2-x) \), the outer function is \( \sin(u) \) and the inner function is \( u = 2-x \). The chain rule states that the derivative of the composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

In mathematical terms, if a function \( y = f(g(x)) \), then the derivative \( y' \) is \( f'(g(x)) \cdot g'(x) \). This approach allows us to differentiate complex expressions effectively as demonstrated in the exercise.

Composite Functions

Composite functions are functions within functions. These are created by combining two or more functions. The key to recognizing composite functions is identifying that you have one function inside another. In our example, \( f(x) = \sin(2-x) \), the outer function is \( \sin(u) \) and the inner function is \( u = 2-x \).
The beauty of composite functions lies in how they allow for more complex relationships and transformations in mathematical expressions. When dealing with them, it's important to clearly identify which function is the outer and which is the inner,
because this identification is crucial when differentiating using techniques like the chain rule.

• Start by identifying the outer and inner functions.
• Next, apply the chain rule to differentiate each component correctly.

Mastering composite functions is a vital part of calculus that enables you to handle nested and layered expressions with ease.

Trigonometric Differentiation

Trigonometric differentiation is the process of finding the derivative of trigonometric functions like sine, cosine, and tangent. These functions are prevalent in various analyses, especially in periodic phenomena and oscillations.

In trigonometric differentiation, each trigonometric function follows specific rules:

• The derivative of \( \sin(x) \) is \( \cos(x) \).
• The derivative of \( \cos(x) \) is \( -\sin(x) \).
• The derivative of \( \tan(x) \) is \( \sec^2(x) \).

When mixed with other calculus concepts like the chain rule, trigonometric differentiation allows for analyzing composite functions that include trigonometric elements.

In the example provided, \( f(x) = \sin(2-x) \), we differentiated \( \sin(u) \) with respect to \( u \) as \( \cos(u) \), combining this with the chain rule allowed us to find the overall derivative easily.