Problem 45 - Calculus of a Single Variable

Question

Finding a Derivative In Exercises \(37-58\) , find the derivative of the function. (Hint: In some exercises, you may find it helpful to apply logarithmic properties before differentiating.) $$ h(\theta)=2^{-\theta} \cos \pi \theta $$

Step-by-Step Solution

Answer

The derivative of the function is \(h'(\theta) = 2^{-\theta} \cos (\pi \theta) (-\ln 2 - \pi \tan(\pi \theta))\).

Step 1: Convert the given function

We have to convert the function into a form that would make it easier to apply the derivative rules. This can be achieved by taking the natural logarithm on both sides, while making use of the logarithmic properties (mainly power out). So, we have \(\ln h(\theta) = \ln (2^{-\theta} \cos (\pi \theta)) = -\theta \ln 2 + \ln (\cos (\pi \theta))\)

Step 2: Differentiate on both sides

Let's apply the chain rule and the derivative of logarithmic and trigonometric functions to differentiate. Thus we have \(\frac{h'(\theta)}{h(\theta)} = -\ln 2 - \pi \tan(\pi \theta)\)

Step 3: Solve for \(h'(\theta)\)

Finally, we have to solve for \(h'(\theta)\) by multiplying both sides by \(h(\theta)\). So, \(h'(\theta) = h(\theta) (-\ln 2 - \pi \tan(\pi \theta)) = 2^{-\theta} \cos (\pi \theta) (-\ln 2 - \pi \tan(\pi \theta))\).

Key Concepts

Derivatives

Derivatives are a fundamental concept in calculus, representing the rate at which a function changes at any given point. It is essentially the slope of the tangent line to the function's graph at a particular point. For a given function \( f(x) \), the derivative \( f'(x) \) is defined as the limit:\[\lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h}\]This expression gives us the best linear approximation of the function near that point. Derivatives can help us understand how a function behaves with respect to changes in its variables, making them incredibly useful in fields such as physics, engineering, and economics.
Some common rules for finding derivatives include the power rule, product rule, quotient rule, and chain rule, each designed to simplify the differentiation process depending on the form of the function.

Chain Rule

The chain rule is a key derivative rule used when dealing with composite functions, which are functions nested inside one another. If you have a function \( y = f(g(x)) \), then the chain rule states that the derivative \( \frac{dy}{dx} \) is:\[\frac{dy}{dx} = f'(g(x)) \cdot g'(x) \]This rule allows us to differentiate the outer function while respecting how the inner function changes, effectively 'chaining' these derivatives together.

• Start by identifying the outer and inner functions.
• Differentiate the outer function with respect to the inner function.
• Multiply by the derivative of the inner function.

In the problem \( h(\theta) = 2^{-\theta} \cos(\pi \theta) \), the chain rule helps in differentiating each composite part—\(2^{-\theta}\) involves \(-\theta\) and \(\cos(\pi \theta)\) involves \(\pi \theta\), both requiring separate applications of the chain rule.

Logarithmic Differentiation

Logarithmic differentiation is a useful technique to find the derivatives of complex functions, especially those involving products, quotients, or powers. By taking the logarithm of both sides of an equation \( y = f(x) \), it simplifies multiplication into addition:\[\ln y = \ln f(x)\]The derivative can then be found using the properties of logarithms, where the derivative of \(\ln y\) becomes \( \frac{y'}{y} \). Multiply through by \( y \), to find \( y' \).
This method is especially handy for differentiating expressions like \( h(\theta) = 2^{-\theta} \cos(\pi \theta) \).

• Use \( \ln(a \cdot b) = \ln a + \ln b \) to separate the function.
• Apply derivatives to each part using standard rules.

This technique is integral for simplifying otherwise complex differentiation tasks.

Trigonometric Functions

Trigonometric functions like \( \sin \theta \), \( \cos \theta \), and \( \tan \theta \) are functions of an angle and are fundamental in studying periodic phenomena. Each has its derivative:

• The derivative of \( \sin x \) is \( \cos x \).
• The derivative of \( \cos x \) is \(-\sin x \).
• The derivative of \( \tan x \) is \( \sec^2 x \).

In the exercise, the trigonometric function \( \cos(\pi \theta) \) appears. Its derivative is found using the chain rule, as the function is composed with a linear transformation \( \pi \theta \). Calculating \[-\pi \sin(\pi \theta)\], this term is crucial in determining how the trigonometric aspect contributes to the function's rate of change. Understanding these derivative relationships is vital in any calculus problem involving periodic or oscillatory behavior.