Problem 70 - Calculus for Biology and Medicine

Question

Find the derivatives of the following functions: $$ f(x)=\cos ^{2}\left(2 x^{2}+3\right) $$

Step-by-Step Solution

Answer

The derivative is \( \frac{df}{dx} = -8x \cos(2x^2 + 3) \sin(2x^2 + 3) \).

Step 1: Identify the Structure

The given function is \( f(x) = \cos^2(2x^2 + 3) \). It is a composite function, where the outer function is \( u^2 \) and the inner function is \( u = \cos(v) \), where \( v = 2x^2 + 3 \). We will need to use the chain rule to find the derivative.

Step 2: Differentiate the Outer Function

The outer function is \( u^2 \), where \( u = \cos(v) \). The derivative of \( u^2 \) with respect to \( u \) is \( 2u \). The derivative of the outer function with respect to \( u \) is: \[ \frac{d}{du}(u^2) = 2u. \]

Step 3: Differentiate the Middle Function

Now consider the middle function \( u = \cos(v) \), where \( v = 2x^2 + 3 \). The derivative of \( \cos(v) \) with respect to \( v \) is \( -\sin(v) \).

Step 4: Differentiate the Inner Function

The inner function is \( v = 2x^2 + 3 \). The derivative of \( 2x^2 + 3 \) with respect to \( x \) is \( 4x \).

Step 5: Apply the Chain Rule

Using the chain rule to combine the derivatives: \[ \frac{df}{dx} = \frac{d}{dx}(\cos^2(2x^2 + 3)) = \frac{d}{du}(u^2) \cdot \frac{du}{dv}(\cos(v)) \cdot \frac{dv}{dx}(2x^2 + 3). \]Substituting back the derivatives computed:\[ \frac{df}{dx} = 2(\cos(v)) \cdot (-\sin(v)) \cdot (4x). \]Replace \( v \) with \( 2x^2 + 3 \):\[ \frac{df}{dx} = 2(\cos(2x^2 + 3))(-\sin(2x^2 + 3))(4x). \]

Step 6: Simplify the Expression

Simplify the expression for the derivative:\[ \frac{df}{dx} = -8x(\cos(2x^2 + 3))(\sin(2x^2 + 3)). \]

Key Concepts

Derivatives

A derivative represents the rate of change of a function with respect to a variable. In simple terms, it tells you how a function's output changes as its input changes. The derivative is a fundamental concept in calculus and is denoted by \( f'(x) \) or \( \frac{df}{dx} \) when considering the variable \( x \). For a standard function like \( x^n \), the derivative is calculated using the power rule, resulting in \( nx^{n-1} \). This is rather straightforward when dealing with basic functions.

However, in more complex situations, such as when dealing with composite functions or when multiple functions are combined, different rules like the product rule, quotient rule, and chain rule are utilized. The process of finding a derivative is often called differentiation.

• Power Rule: \( \frac{d}{dx} (x^n) = nx^{n-1} \)
• Chain Rule: Used for functions within functions, like \( f(g(x)) \).

Understanding derivatives is crucial because they allow us to predict future behavior based on current data, which is applicable in various disciplines such as physics, engineering, and economics.

Composite Function

A composite function combines two or more functions where the output of one function becomes the input for another. Imagine a function \( f(x) \) and another function \( g(x) \). A composite function \( f(g(x)) \) processes \( x \) through \( g(x) \) first, then applies \( f(x) \) on the resulting value. This layering of functions is common in complex mathematical problems.

In this exercise, the function \( \cos^2(2x^2 + 3) \) is composite because it consists of the outer function \( u^2 \), where \( u = \cos(v) \), and an inner function \( v = 2x^2 + 3 \). The key to working with composite functions is understanding how to differentiate them effectively using the chain rule.

• Structure of Composite Functions: They are written \( f(g(x)) \), showing a hierarchy of operations.
• Chain Rule: Essential for differentiating composite functions. It combines derivatives in a specific way.

Trigonometric Functions

Trigonometric functions, such as sine, cosine, and tangent, are foundational in both geometry and calculus. They originate from the relationships between the angles and sides of triangles, especially right triangles, and are extended to describe waves and oscillatory motions. Each function has distinctive properties and derivatives which are crucial in calculus.

For example, the derivative of the cosine function, \( \cos(x) \), is \(-\sin(x) \). This negative sign indicates the function's rate and direction of change, a pattern essential in more complex derivatives, especially when combined with the chain rule.

• Key Trigonometric Derivatives:
  • \( \frac{d}{dx} \cos(x) = -\sin(x) \)
  • \( \frac{d}{dx} \sin(x) = \cos(x) \)
• Applications: Trigonometric derivatives are vital in analyzing periodic functions and waveforms in physics and engineering.

In the provided exercise, the function involves \( \cos(2x^2 + 3) \), and recognizing how to differentiate this using the chain rule is crucial to solving the derivative of the entire expression.