Problem 28 - Thomas Calculus

Question

In Exercises \(17-36,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$ y=\theta^{3} e^{-2 \theta} \cos 5 \theta $$

Step-by-Step Solution

Answer

The derivative is \((3\theta^2 - 2\theta^3) e^{-2\theta} \cos 5\theta - 5\theta^3 e^{-2\theta} \sin 5\theta\)."

Step 1: Identify the Rule

The function is a product of three functions: \(y = u \cdot v \cdot w\) where \(u = \theta^3\), \(v = e^{-2\theta}\), and \(w = \cos 5\theta\). To find the derivative, we will apply the product rule for three functions: \( \frac{d}{d\theta}(u \cdot v \cdot w) = u' \cdot v \cdot w + u \cdot v' \cdot w + u \cdot v \cdot w' \).

Step 2: Differentiate Individual Parts

Calculate the derivatives of \(u\), \(v\), and \(w\) with respect to \(\theta\):1. \(u' = \frac{d}{d\theta}(\theta^3) = 3\theta^2\).2. \(v' = \frac{d}{d\theta}(e^{-2\theta}) = -2e^{-2\theta}\).3. \(w' = \frac{d}{d\theta}(\cos 5\theta) = -5\sin 5\theta\).

Step 3: Apply the Product Rule for Derivatives

Using the product rule for three functions:\[\frac{d}{d\theta}(\theta^3 e^{-2\theta} \cos 5\theta) = (3\theta^2)(e^{-2\theta})(\cos 5\theta) + (\theta^3)(-2e^{-2\theta})(\cos 5\theta) + (\theta^3)(e^{-2\theta})(-5\sin 5\theta)\]Simplify each term:- First term: \(3\theta^2 e^{-2\theta} \cos 5\theta\).- Second term: \(-2\theta^3 e^{-2\theta} \cos 5\theta\).- Third term: \(-5\theta^3 e^{-2\theta} \sin 5\theta\).

Step 4: Combine and Simplify

Combine like terms to simplify the expression:\[\frac{dy}{d\theta} = 3\theta^2 e^{-2\theta} \cos 5\theta - 2\theta^3 e^{-2\theta} \cos 5\theta - 5\theta^3 e^{-2\theta} \sin 5\theta\]Further simplifying the first two terms:\[\frac{dy}{d\theta} = (3\theta^2 - 2\theta^3) e^{-2\theta} \cos 5\theta - 5\theta^3 e^{-2\theta} \sin 5\theta\]The final derivative is:\[\frac{dy}{d\theta} = (3\theta^2 - 2\theta^3) e^{-2\theta} \cos 5\theta - 5\theta^3 e^{-2\theta} \sin 5\theta\]

Step 5: Verify the Solution

Verify each derivative step and simplification to ensure correctness. This involves checking each differentiation and the final product rule application for accuracy.

Key Concepts

product rule

The product rule is an essential tool in calculus, particularly when dealing with the derivatives of products of functions. Imagine you have functions multiplied together, like in this exercise where we have three functions: \( u = \theta^3 \), \( v = e^{-2\theta} \), and \( w = \cos 5\theta \). When differentiating a product of functions, simply multiplying their derivatives will not suffice. Instead, the product rule must be applied.

For two functions, the product rule states:

• \( \frac{d}{dx}(uv) = u'v + uv' \)

With three functions, the rule extends as follows:

• \( \frac{d}{d\theta}(uvw) = u'vw + uv'w + uvw' \)

Each part of the original function is differentiated individually while keeping the other parts intact. This ensures you've accounted for how each function contributes to the overall change when the variable changes. Learning and applying the product rule enables you to tackle complex differentiation problems efficiently.

differentiation

Differentiation is the process of finding the derivative of a function, which represents the rate of change. In this exercise, we find the derivative of a composite function \( y = \theta^3 e^{-2\theta} \cos 5\theta \) with respect to \( \theta \).

Let's break down the differentiation of each component:

• \( u = \theta^3 \) differentiates to \( u' = 3\theta^2 \).
• \( v = e^{-2\theta} \) differentiates to \( v' = -2e^{-2\theta} \).
• \( w = \cos 5\theta \) differentiates to \( w' = -5\sin 5\theta \).

After differentiating each part, the derivatives are substituted into the product rule formula (specifically for three products). Each derivative tells us how that part of the function changes with respect to \( \theta \), contributing to the overall change in \( y \).Differentiation requires careful handling of the rules, especially as functions become more complex with multiple layers of operations involved.

multivariable functions

Multivariable functions introduce even more layers to differentiation, as they involve more than one independent variable. In this exercise, we focus on a single variable \( \theta \), but understanding multivariable functions is crucial for more advanced calculus problems.

In multivariable calculus, we often deal with partial derivatives, which involve differentiating with respect to one variable while treating others as constants. However, when dealing with a product of several functions of the same variable, like in our problem, the product rule becomes a vital technique.

• For example, if a function was instead \( f(x, y) = xy\theta \), you would have to decide which variable you differentiate concerning \( x \) or \( y \) first.

When transitioning from single variables to multivariable functions, it becomes more important to fully grasp the product rule and differentiation as many real-world phenomena are explained through multivariable models. By practicing with simpler constructs, like in this exercise, understanding complex functions in multivariable settings becomes much more manageable.