Problem 18 - Calculus for Biology and Medicine

Question

Find the derivative with respect to the independent variable. $$ f(x)=\cos ^{2}\left(x^{2}-1\right) $$

Step-by-Step Solution

Answer

The derivative is \(-2x\sin(2x^2-2)\).

Step 1: Understand the Function

Identify that the function is a composition, specifically a squared trigonometric function. We have \( f(x)=\cos ^{2}ig(x^{2}-1\big) \), which complicates the differentiation because it involves both a chain rule for the outer function and an additional chain rule for the argument of the cosine function.

Step 2: Apply the Chain Rule - Outer Function

Consider \( u = \cos(x^2 - 1) \), thus \( f(x) = u^2 \). Using the chain rule, the derivative of \( f(x) = u^2 \) with respect to \( x \) is \( 2u \times \frac{du}{dx} \). We'll calculate \( \frac{du}{dx} \) in the next step.

Step 3: Differentiate the Inner Function

Differentiate \( u = \cos(x^2 - 1) \) with respect to \( x \). This requires the chain rule where \( \frac{du}{dx} = \frac{d}{dx}[\cos(x^2-1)] = -\sin(x^2-1) \times \frac{d}{dx}[x^2 - 1] = -\sin(x^2-1) \times 2x \).

Step 4: Combine Results

Substitute back \( u = \cos(x^2-1) \) and \( \frac{du}{dx} = -2x\sin(x^2-1) \) into the expression for \( \frac{df}{dx} \). Thus, the derivative is \( \frac{df}{dx} = 2\cos(x^2-1)[-2x\sin(x^2-1)] \).

Step 5: Simplify the Expression

Simplify \( \frac{df}{dx} = -4x\cos(x^2-1)\sin(x^2-1) \). Recognize that this is equivalent to \( -2x\sin(2(x^2-1)) \) using the double-angle identity \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \).

Key Concepts

Understanding Derivatives

Derivatives are a fundamental concept in calculus. They measure how a function changes as its input changes. In simpler terms, a derivative tells us the rate at which a function is increasing or decreasing at any given point.
A derivative is often written as \( f'(x) \) or \( \frac{df}{dx} \), representing the change in \( f \) with respect to \( x \). This concept is crucial for analyzing the behavior of functions in various contexts, such as physics for determining velocity or in economics to find marginal costs.
When dealing with complex functions, especially those involving compositions like trigonometric functions, finding derivatives requires specific rules and techniques. This is where the chain rule comes in, which is essential when a function is composed of other functions. It's like peeling layers off an onion, taking one derivative at each step. This technique allows us to handle derivatives of nested functions efficiently.

Trigonometric Functions and Their Derivatives

Trigonometric functions, such as sine and cosine, play significant roles in calculus and are used to describe oscillatory behavior. Understanding how to differentiate these functions is key in calculus.
The basic derivatives you need to know are:

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

Using these relationships allows us to differentiate more complex functions containing trigonometric expressions. When a trigonometric function is composed with another function, as in our example with \( \, \cos^2(x^2-1) \, \), the chain rule becomes specifically useful.
These trigonometric identities and their derivatives form a base that helps us simplify and solve a wide array of problems in calculus, especially those involving periodic phenomena.

Understanding the Composition of Functions

A composition of functions is when one function is placed inside another. Consider a function \( f(x) \) inside another function \( g(x) \), making a new function \( h(x) = g(f(x)) \). This is written as \( h(x) = g(f(x)) \). The powerful chain rule allows us to find the derivative of these compositions easily.
The chain rule states: \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \). This means we differentiate the outer function (\( f \)), keeping the inner function (\( g \)) as it is, and then multiply by the derivative of the inner function. In our example, we looked at the function \( f(x) = u^2 \), where \( u = \cos(x^2 - 1) \).
Understanding compositions is crucial because many real-world situations involve such nested functions. Proficiency with this helps in unraveling complicated derivatives, leading to accurate calculations related to rates of change, optimization problems, and much more. It's like unlocking a multistep puzzle, with each function revealing another layer to explore.