Problem 22 - Calculus for Biology and Medicine

Question

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

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

Step 1: Identify the parts of the function

The given function is composed of two main parts: \( -5 \cos(2-x^3) \) and \( 2 \cos^3(x-4) \). Each part needs to be derived separately, and we'll apply the chain rule for each component.

Step 2: Differentiate \(-5 \cos(2-x^3)\)

First, differentiate \(-5 \cos(2-x^3)\). The derivative of \( \cos(u) \) with respect to \( u \) is \( -\sin(u) \). Then, apply the chain rule: \[\frac{d}{dx}[-5 \cos(2-x^3)] = -5 \cdot [-\sin(2-x^3)] \cdot \frac{d}{dx}(2-x^3) \] Since \( \frac{d}{dx}(2-x^3) = -3x^2 \), the derivative becomes: \[5 \sin(2-x^3)(-3x^2) = -15x^2 \sin(2-x^3)\]

Step 3: Differentiate \(2 \cos^3(x-4)\)

Next, differentiate \(2 \cos^3(x-4)\). Use the chain rule on the cubed cosine function: \[ \frac{d}{dx}[2 \cos^3(x-4)] = 2 \cdot 3 \cos^2(x-4) \cdot \frac{d}{dx}[\cos(x-4)] \]The derivative of \(\cos(x-4)\) is \(-\sin(x-4)\), resulting in: \[2 \cdot 3 \cos^2(x-4) \cdot (-\sin(x-4)) \]Simplify the expression to: \[-6 \cos^2(x-4) \sin(x-4)\]

Step 4: Combine derivatives

Combine the derivatives from Step 2 and Step 3 to get the overall derivative of the function \( f(x) \):\[f'(x) = -15x^2 \sin(2-x^3) - 6 \cos^2(x-4) \sin(x-4)\]

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus, especially useful when dealing with composite functions. It enables us to differentiate a function composed of multiple functions, by breaking it down into simpler parts. Imagine you have a function within a function, like an onion with layers. The chain rule "peels" these layers.

• First, differentiate the outer function, while keeping the inner function unchanged.
• Next, differentiate the inner function.
• Finally, multiply the derivatives from these two steps.

This approach is akin to systematically zooming in on each function level, ensuring no derivative detail is overlooked.

Consider the term \(-5 \cos(2-x^3)\). Here, \(\cos(u)\) is outside \((2-x^3)\), our inside function. We start by differentiating the outer function \(-5 \cos(u)\), giving us \(5 \sin(u)\). Then, we differentiate the inner function \(2-x^3\), resulting in a derivative of \(-3x^2\). Multiply the derivative of the outer and inner functions together for the final result: \(-15x^2 \sin(2-x^3)\).

This method applies uniformly to complex multi-layered functions, which makes the chain rule indispensable in calculus.

Trigonometric Differentiation

Trigonometric differentiation specifically deals with functions involving trigonometric expressions like sine, cosine, and tangent. The trigonometric rules complement the power rule, product rule, and especially the chain rule.

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

These basic rules allow us to handle more complex expressions that incorporate trigonometric functions.

In the exercise, \(-5 \cos(2-x^3)\) invokes trigonometric differentiation; we first apply the basic derivative of cosine, resulting in \(-5 \times -\sin(2-x^3)\). Similarly, for \(2 \cos^3(x-4)\), note that the cube exponent means we are dealing with \(\cos(u)^3\), requiring differentiation twice: firstly, by using the power rule, and then by applying trigonometric differentiation for \(\cos(x-4)\). This yielded \(-6 \cos^2(x-4) \sin(x-4)\).

So, trigonometric differentiation helps dissect the sine and cosine elements into more manageable mathematical expressions.

Function Composition

Function composition occurs when one function is nested within another, like stacking blocks. It's an essential part of understanding how the chain rule is applied since it involves differentiating these nested relationships.

• A composed function often looks like \(f(g(x))\), where \(g(x)\) is the inside or inner function.
• When differentiating, the change in one function affects the other, which is why the chain rule is used.

This nested arrangement means we first address the outer function and then tackle the inner one.

In this exercise, both parts of the function \(-5 \cos(2-x^3)\) and \(2 \cos^3(x-4)\) demonstrated function composition. The cosine functions have inner compositions \(2-x^3\) and \((x-4)\) respectively. To find the derivative, recognizing this layout enables us to apply the chain rule effectively for evaluating the differential impact of each component.

Understanding function composition is thus key to unraveling the nested structure of more complex calculus problems, making it simpler to apply the correct differentiation rules.