Problem 23 - Calculus for Biology and Medicine

Question

Find the derivative with respect to the independent variable. $$ f(x)=4 \cos ^{2} x+2 \cos x^{4} $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = -8 \cos(x) \sin(x) - 8x^3 \sin(x^4). \)

Step 1: Understand the Function

The given function is a combination of trigonometric and algebraic functions: \[ f(x) = 4 \cos^2(x) + 2 \cos(x^4). \]We need to find the derivative with respect to the variable \( x \).

Step 2: Apply the Chain Rule

To differentiate \( \cos^2(x) \), use the chain rule. The chain rule states \( (g(h(x)))' = g'(h(x)) \cdot h'(x) \). Here, let \( u = \cos(x) \), then \( f(x) = 4u^2 \). The derivative is \( \frac{d}{dx} 4u^2 = 8u \cdot \frac{du}{dx} = 8\cos(x) \cdot (-\sin(x)) = -8\cos(x)\sin(x). \)

Step 3: Differentiate Second Term

The second term is \( 2 \cos(x^4) \). Again, apply the chain rule:1. The outer function \( g(u) = \cos(u) \) has derivative \( -\sin(u) \).2. Let \( v = x^4 \), then \( u = v \) and \( h(v) = \cos(v) \).3. The derivative of \( v = x^4 \) is \( 4x^3 \).Hence, the derivative is \(-2 \sin(x^4) \cdot 4x^3 = -8x^3\sin(x^4). \)

Step 4: Combine the Derivatives

Add the derivatives of both terms to find the derivative of the function: \[ f'(x) = -8 \cos(x) \sin(x) - 8x^3 \sin(x^4). \]

Step 5: Simplify, if Possible

The derivative can be left in the current format. It is simplified and ready:\[ f'(x) = -8 \cos(x) \sin(x) - 8x^3 \sin(x^4). \]Double-check calculations to ensure accuracy.

Key Concepts

Understanding the Chain Rule in Differentiation

One of the most powerful tools in calculus is the chain rule. It helps us find derivatives of composite functions. In simpler terms, whenever we have a function inside another function, the chain rule comes into play.
For example, if we have a function like \( g(h(x)) \), the chain rule states:

• First, find the derivative of the outer function \( g \) in terms of the inner function \( h(x) \).
• Then, multiply by the derivative of the inner function \( h(x) \).

So, the derivative \((g(h(x)))' = g'(h(x)) \cdot h'(x)\).
In our exercise, for the term \( \cos^2(x) \), let \( u = \cos(x) \). Though it initially seems like a single function, the internal component \( \cos(x) \) means we will use the chain rule twice: first for the power and then for cosine itself. The same method applies to \( \cos(x^4) \), where an additional inner layer \( x^4 \) is involved.
By carefully applying the chain rule to each layer, these derivatives turn into neatly solvable pieces.

Exploring Trigonometric Differentiation

Trigonometric functions like cosine and sine have their own rules for differentiation. These functions are commonplace in calculus and require different techniques compared to polynomial functions.
Consider, for instance, the standard derivatives:

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

When these trigonometric functions are part of a larger chain, as in this exercise, their derivatives mesh with the algebraic forms through chain rule applications. The term \( \cos^2(x) \) involves both power and trigonometric differentiation, which makes it a bit more complex.
Once you grasp the derivatives of basic trigonometric functions and know when to use the chain rule, tackling more complicated expressions becomes much easier. In the problem at hand, recognizing the difference between a simple trigonometric derivative and one that requires additional chain rule unlatches easier understanding.

Algebraic Functions and Their Derivatives

Algebraic functions, such as polynomials, have derivatives that often follow straightforward rules. The power rule is the most common method used, where the derivative of \( x^n \) is \( nx^{n-1} \).
In this specific exercise, the term \( x^4 \) embedded within the cosine function needs differential handling through the chain rule. Breaking it into pieces, one finds the derivative of \( x^4 \) as an isolated algebraic function first, then merges it with the trigonometric part.
This combination underscores a fundamental strength of calculus: the ability to approach intricate functions by part-separating them into algebraic components and overlapping principles. Recognizing these separate rules and their applicability builds a foundation for tackling various differentiable functions.