Problem 17 - Thomas Calculus

Question

Find the derivative of \(y\) with respect to the given independent variable. \(y=(\cos \theta)^{\sqrt{2}}\)

Step-by-Step Solution

Answer

The derivative is \( -\\sqrt{2} (\\cos \theta)^{\\sqrt{2} - 1} \\sin \theta \).

Step 1: Understand the Function

The function given is \( y = (\cos \theta)^{\sqrt{2}} \). This is a composite function where the inner function is \( \cos \theta \), and the outer function is taking that result to the power of \( \sqrt{2} \). We will need the chain rule to find the derivative.

Step 2: Apply the Chain Rule

The chain rule states that the derivative of \( y = f(g(x)) \) with respect to \( x \) is \( f'(g(x)) \cdot g'(x) \). In this problem, \( f(u) = u^{\sqrt{2}} \) and \( g(\theta) = \cos \theta \). We need to find \( f'(u) \) and \( g'(\theta) \).

Step 3: Find the Derivative of the Outer Function

Let \( u = \cos \theta \). Then \( f(u) = u^{\sqrt{2}} \), and the derivative \( f'(u) = \sqrt{2}u^{\sqrt{2}-1} \).

Step 4: Find the Derivative of the Inner Function

For the inner function \( g(\theta) = \cos \theta \), the derivative \( g'(\theta) = -\sin \theta \).

Step 5: Combine Using the Chain Rule

Substitute into the chain rule: \(\frac{dy}{d\theta} = f'(g(\theta)) \cdot g'(\theta) = \sqrt{2} (\cos \theta)^{\sqrt{2} - 1} \cdot (-\sin \theta) \).

Step 6: Simplify the Expression

Combine the terms: \(\frac{dy}{d\theta} = -\sqrt{2} (\cos \theta)^{\sqrt{2} - 1} \sin \theta \). This is the derivative of \( y \) with respect to \( \theta \).

Key Concepts

Chain Rule

The chain rule is an essential tool in calculus used for finding the derivative of composite functions. Composite functions involve two or more functions composed together, where one function is inside another. In simple terms, if you have a function like \( y = f(g(x)) \), the chain rule helps you differentiate it effectively. The rule states:

• The derivative \( \frac{dy}{dx} \) is obtained by multiplying the derivative of the outer function \( f'(g(x)) \) with the derivative of the inner function \( g'(x) \).

Consider an example where the outer function is \( f(u) = u^{\sqrt{2}} \) and the inner function is \( u = \cos \theta \). When applying the chain rule, you first differentiate the outer function with respect to \( u \) and then differentiate the inner function with respect to \( \theta \). Together, these derivatives get multiplied to give you the final derivative. This step-by-step multiplication lies at the heart of the chain rule application.

Trigonometric Functions

Trigonometric functions like sine and cosine are fundamental in calculus due to their periodic nature and wide applicability. In this exercise, you encounter the cosine function, expressed as \( \cos \theta \).

• These functions are based on the unit circle and they describe the relationship between the angles and side lengths of triangles.
• Understanding how to differentiate these functions is crucial for solving calculus problems involving oscillations or waves.

The derivative of the cosine function is the negative sine function, \( \frac{d}{d\theta} \cos \theta = -\sin \theta \). This relationship shows that as the angle changes, how quickly the cosine function decreases is determined by the sine function's value at that angle. It's important to remember this when applying the chain rule to composite functions involving trigonometric terms.

Power Rule

The power rule is a fundamental differentiation rule that provides an easy way to find the derivative of power functions. If you have a function \( f(x) = x^n \), the power rule states that the derivative is \( f'(x) = n \cdot x^{n-1} \). This rule applies when the function is expressed in terms of a simple power of the variable.

• The power rule simplifies the process of differentiation when combined with other rules, like the chain rule.
• In cases where the exponent isn’t an integer or involves other expressions, the rule still applies by handling those components separately.

In our original problem, \( f(u) = u^{\sqrt{2}} \), the power rule helps us find the derivative \( f'(u) = \sqrt{2} \cdot u^{\sqrt{2} - 1} \). By systematically applying the power rule, complex expressions become manageable, making them a perfect complement to the chain rule when dealing with powers of both simple and complex expressions.