Problem 16 - Calculus for Biology and Medicine

Question

In Problems \(1-58\), find the derivative with respect to the independent variable. $$ f(x)=2 \cos \left(x^{3}-3 x\right) $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = -6x^2 \sin(x^3 - 3x) + 6 \sin(x^3 - 3x) \).

Step 1: Identify the function structure

The given function is written as \( f(x) = 2 \cos(x^3 - 3x) \). The function has an outer function, the cosine, and an inner function, \( g(x) = x^3 - 3x \). This indicates that we will use the chain rule to differentiate it.

Step 2: Differentiate the outer function

Start by differentiating the outer function, which is \( 2 \cos(u) \), where \( u = x^3 - 3x \). The derivative of \( \cos(u) \) is \( -\sin(u) \). Thus, the derivative of the outer function is \( -2 \sin(u) \) with respect to \( u \).

Step 3: Differentiate the inner function

Next, find the derivative of the inner function \( g(x) = x^3 - 3x \). Differentiating term by term, we obtain \( g'(x) = 3x^2 - 3 \).

Step 4: Apply the chain rule

Use the chain rule, which states that the derivative of \( f(x) = 2 \cos(g(x)) \) is \( f'(x) = 2 \cdot (-\sin(g(x))) \cdot g'(x) \). Substitute \( g(x) = x^3 - 3x \) and \( g'(x) = 3x^2 - 3 \). This results in the derivative: \[ f'(x) = -2 \sin(x^3 - 3x) \cdot (3x^2 - 3) \].

Step 5: Simplify the expression

Distribute to simplify the result: \[ f'(x) = -2(3x^2 - 3) \sin(x^3 - 3x) = -6x^2 \sin(x^3 - 3x) + 6 \sin(x^3 - 3x) \].

Key Concepts

Chain Rule

The chain rule is a powerful technique used in differentiation. It is essential when dealing with composite functions, where one function is nested inside another. In the exercise we're discussing, the chain rule allows us to differentiate the function

• Outer function: \(2 \cos(u)\)
• Inner function: \(u = x^3 - 3x\)

The chain rule tells us how to differentiate such functions. It simplifies this to the formula:\[ f'(x) = f'(g(x)) \cdot g'(x) \]
This means 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 itself. By applying the chain rule correctly, we can handle even the most complex functions with ease.

Trigonometric Functions

Trigonometric functions like sine and cosine are fundamental in calculus. Their derivatives are tools we frequently apply, especially when dealing with oscillatory motion, waves, and circles. In our problem, we specifically look at the cosine function, \(\cos(u)\).
The derivative of the cosine function, \(\cos(u)\), is \(-\sin(u)\). This derivative is crucial when applying the chain rule to problems involving trigonometric identities. In our case:

• Differentiate the outer function \(2 \cos(u)\) as \(-2 \sin(u)\).

Understanding how to differentiate trigonometric functions is key, not only in calculus but also in physical applications, where they often describe periodic phenomena.

Differentiation Techniques

Differentiation is the process of finding the derivative or rate of change of a function. Several techniques exist to simplify this process:

• Product Rule: Used when you have two functions multiplied together.
• Quotient Rule: Used for functions divided by each other.
• Chain Rule: Used for composite functions, as discussed here.

In the exercise, we utilized the chain rule, which is a cornerstone in handling functions within functions. Additionally, we used basic differentiation rules to find the derivative of a polynomial, \(x^3 - 3x\), resulting in \(3x^2 - 3\).
Mastering these techniques simplifies differentiating complex expressions. By combining methods like term-by-term differentiation and the chain rule, we can systematically solve for derivatives in multifaceted problems. Such skills are indispensable for tackling calculus challenges across various scientific and engineering disciplines.