Problem 22 - Calculus for Biology and Medicine

Question

In Problems \(1-58\), find the derivative with respect to the independent variable. $$ f(x)=-5 \cos \left(2-x^{3}\right)+2 \cos ^{3}(x-4) $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = -15x^2 \sin(2-x^3) - 6 \cos^2(x-4) \sin(x-4) \).

Step 1: Understanding the Problem

We need to find the derivative of the function \( f(x) = -5 \cos(2-x^3) + 2 \cos^3(x-4) \) with respect to \( x \). This involves differentiating two separate terms involving the cosine function.

Step 2: Differentiating the First Term

The first term is \(-5 \cos(2-x^3)\). Apply the chain rule: the derivative of \(-\cos(u)\) is \( \sin(u)\) times the derivative of \(u\) which is \(-1\), so the derivative becomes \(5 \sin(2-x^3)\). Now, differentiate the inner function \(2-x^3\) with respect to \(x\), which yields \(-3x^2\). Therefore, the derivative of the first term is \(5 \cdot \sin(2-x^3) \cdot (-3x^2) = -15x^2 \sin(2-x^3)\).

Step 3: Differentiating the Second Term

The second term is \(2 \cos^3(x-4)\). Again, use the chain rule: treat \( \cos(x-4) \) as \(u\). The derivative of \( u^3 \) with respect to \( u \) is \( 3u^2 \), thus the derivative of \( \cos^3(x-4) \) is \( 3 \cos^2(x-4) \cdot (-\sin(x-4))\) times the derivative of \(x-4\), which is \(1\). Therefore, the derivative of the term is \(2 \cdot 3 \cdot \cos^2(x-4) \cdot (-\sin(x-4)) = -6 \cos^2(x-4) \sin(x-4)\).

Step 4: Combining the Results

Add the derivatives of each term to get the final result. Thus, the derivative \( f'(x) \) is: \[ f'(x) = -15x^2 \sin(2-x^3) - 6 \cos^2(x-4) \sin(x-4) \]

Key Concepts

Chain Rule

The chain rule is essential when differentiating composite functions. In simple terms, it is used when you have a function inside another function. To differentiate such a function, you follow a two-step process:

• Differentiating the outer function while keeping the inner function unchanged.
• Multiplying the result by the derivative of the inner function.

Let's consider the function \( -5 \cos(2-x^3) \). The outer function here is \( -\cos(u) \) and the inner function is \( u = 2-x^3 \).
First, we differentiate \( -\cos(u) \) to get \( \sin(u) \), and since we have a negative sign, it becomes \( \sin(u) \) multiplied by -1.
Next, we differentiate the inner function \( 2-x^3 \) which results in \(-3x^2\).
Finally, combining these derivatives, we get \[ -15x^2 \sin(2-x^3) \] as part of the complete derivative. The chain rule enables us to handle compound functions efficiently by breaking them down into manageable parts.

Trigonometric Functions

Trigonometric functions, like sine and cosine, are commonly encountered in calculus, especially when dealing with periodic behavior or oscillations. Each trigonometric function has its derivative:

• The derivative of \( \sin(x) \) is \( \cos(x) \).
• The derivative of \( \cos(x) \) is \( -\sin(x) \).

These derivatives are essential when performing calculus operations involving trigonometric functions.
In our example, we see both \( \cos(2-x^3) \) and \( \cos(x-4) \). When differentiating, remember that any alteration inside the trigonometric function affects the outcome. For example, in \( \cos(2-x^3) \), the chain rule was applied to accommodate for \(2-x^3\).
Additionally, for functions like \(2 \cos^3(x-4)\), not only the derivative of cosine matters, but also the power of three. Differentiating such composite forms involves using both the chain rule and the power rule. Trigonometric derivatives are crucial when dealing with real-world scenarios such as waveforms and circuits.

Differentiation Techniques

Differentiation is a cornerstone of calculus, providing tools to understand how functions change. Various techniques exist, each useful under different circumstances. The power rule, chain rule, product rule, and quotient rule are among the fundamental methods.
In our problem, the focus is on the chain rule and the combination of trigonometric derivatives. Here's how differentiation was tackled:

• First, identify the different types of terms within the function. Break them down using suitable techniques.
• Apply the appropriate rule to each segment. For instance, for \( -5 \cos(2-x^3) \), use the chain rule.
• Combine all derivative parts to obtain the complete differentiation result.

Being able to apply these techniques seamlessly is vital. They enable you to tackle complex equations effortlessly and gain deeper insights into function behavior. Mastering these differentiation techniques empowers students to solve numerous problems across various fields in mathematics and engineering.