Problem 61 - Thomas Calculus in SI Units

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: Identify the Derivative Rule

The function is of the form \( y = u^n \), where \( u = \cos \theta \) and \( n = \sqrt{2} \). To differentiate \( y \) with respect to \( \theta \), we can use the chain rule in conjunction with the power rule for derivatives.

Step 2: Apply the Power Rule

The derivative of \( y = u^n \) is given by \( \frac{dy}{du} = nu^{n-1} \) where \( u = \cos \theta \) and \( n = \sqrt{2} \). Thus, \( \frac{dy}{du} = \sqrt{2} (\cos \theta)^{\sqrt{2} - 1} \).

Step 3: Differentiate the Inner Function

Now, we need to differentiate the inner function \( u = \cos \theta \) with respect to \( \theta \). The derivative of \( \cos \theta \) is \( -\sin \theta \). Thus, \( \frac{du}{d\theta} = -\sin \theta \).

Step 4: Use the Chain Rule

According to the chain rule, \( \frac{dy}{d\theta} = \frac{dy}{du} \cdot \frac{du}{d\theta} \). Substituting the derivatives we found: \( \frac{dy}{d\theta} = \sqrt{2} (\cos \theta)^{\sqrt{2} - 1} \cdot (-\sin \theta) \).

Step 5: Simplify the Expression

Simplify the expression: \( \frac{dy}{d\theta} = -\sqrt{2} (\cos \theta)^{\sqrt{2} - 1} \sin \theta \).

Key Concepts

Chain Rule

The chain rule is a powerful tool in calculus used when differentiating composite functions. It allows us to find the derivative of a function that is composed of two or more functions. In simpler terms, if you have a function nested inside another function, the chain rule provides a strategy to take derivatives.

Consider a function expressed as \( y = g(f(x)) \). The derivative of this composite function is found by taking the derivative of the outer function evaluated at the inner function, and then multiplying it by the derivative of the inner function. Mathematically, this is written as:

• \( \frac{dy}{dx} = g'(f(x)) \cdot f'(x) \)

Let's take the function \( y = (\cos \theta)^{\sqrt{2}} \). Here, \( u = \cos \theta \) is the inner function, and \( u^{\sqrt{2}} \) is the result of applying the power rule to the outer function. We use the chain rule to find the overall derivative, multiplying the derivative of the outer expression with the derivative of the inner \( \cos \theta \).

Power Rule

The power rule simplifies the process of differentiating functions raised to a power. When you have a base \( u \) raised to some exponent \( n \), the power rule states that the derivative is obtained by multiplying the exponent \( n \) by the base raised to the power of \( n-1 \).

For example, the derivative of \( y = u^n \) with respect to \( u \) is:

• \( \frac{dy}{du} = nu^{n-1} \)

Applying this rule to \( y = (\cos \theta)^{\sqrt{2}} \), we set \( n = \sqrt{2} \) and find the derivative with respect to \( \cos \theta \). It becomes \( \sqrt{2} (\cos \theta)^{\sqrt{2} - 1} \). Afterward, multiplying by the derivative of the inside function (using the chain rule) gives us the total derivative with respect to \( \theta \).

Trigonometric Functions

Trigonometric functions, like sine and cosine, are fundamental in calculus due to their periodic nature. They have specific derivatives that often appear in calculus problems.

The basic derivative of cosine is particularly relevant here:

• \( \frac{d}{d\theta}(\cos \theta) = -\sin \theta \)

This derivative comes into play when applying both the power rule and chain rule. With our original function \( y = (\cos \theta)^{\sqrt{2}} \), once we differentiate the outer function using the power rule, we turn to the cosine function's derivative. This gives a clean final expression for the derivative: \( -\sqrt{2} (\cos \theta)^{\sqrt{2} - 1} \sin \theta \). Understanding these derivatives helps simplify complex differentiation tasks that involve trigonometric functions.