Problem 18 - Thomas Calculus

Question

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. \begin{equation}y=\theta^{3} e^{-2 \theta} \cos 5 \theta\end{equation}

Step-by-Step Solution

Answer

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

Step 1: Identify the Function to Differentiate

We are given the function \(y = \theta^3 e^{-2\theta} \cos(5\theta)\). It involves a product of three functions: \(\theta^3\), \(e^{-2\theta}\), and \(\cos(5\theta)\). We will use the product rule and chain rule to differentiate this expression with respect to \(\theta\).

Step 2: Apply the Product Rule

The product rule states that for three functions \(u\), \(v\), and \(w\), \(\frac{d}{d\theta}(uvw) = uvw' + uv'w + u'vw\). Here, \(u = \theta^3\), \(v = e^{-2\theta}\), \(w = \cos(5\theta)\).

Step 3: Differentiate Each Function

- Differentiate \(u = \theta^3\): \(u' = 3\theta^2\).- Differentiate \(v = e^{-2\theta}\) using the chain rule: \(v' = -2e^{-2\theta}\).- Differentiate \(w = \cos(5\theta)\) using the chain rule: \(w' = -5\sin(5\theta)\).

Step 4: Substitute into the Product Rule

Substitute the differentiated parts into the product rule:\[ \frac{d}{d\theta}(\theta^3 e^{-2\theta} \cos(5\theta)) = \theta^3 e^{-2\theta}(-5\sin(5\theta)) + \theta^3 (-2e^{-2\theta}) \cos(5\theta) + 3\theta^2 e^{-2\theta} \cos(5\theta). \]

Step 5: Simplify the Expression

Combine and simplify the expression:\[ \frac{d}{d\theta}(\theta^3 e^{-2\theta} \cos(5\theta)) = \theta^3 e^{-2\theta}(-5\sin(5\theta)) - 2\theta^3 e^{-2\theta} \cos(5\theta) + 3\theta^2 e^{-2\theta} \cos(5\theta). \]This can be further simplified by factoring out common terms such as \(e^{-2\theta}\).

Step 6: Final Simplification

Factor out \(e^{-2\theta}\) from the simplified expression:\[ \frac{d}{d\theta}(\theta^3 e^{-2\theta} \cos(5\theta)) = e^{-2\theta} \left(-5\theta^3 \sin(5\theta) + \theta^2\cos(5\theta)(3 - 2\theta) \right). \]

Key Concepts

Product Rule

In calculus, the product rule is an essential tool for differentiating functions that are products of two or more terms. This rule helps us find the derivative of a product of functions efficiently. When we have multiple functions being multiplied together, like in our exercise, the product rule allows us to break down the differentiation into manageable parts.

For two functions, say, functions \(f\) and \(g\), the product rule states:

• \(\frac{d}{dx}(f(x) \cdot g(x)) = f(x) g'(x) + g(x) f'(x)\).

When extended to three functions, as in the exercise dealing with \(u = \theta^3\), \(v = e^{-2\theta}\), and \(w = \cos(5\theta)\), the rule becomes:

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

Applying this rule accurately requires differentiating each function individually and summing the results in the specific structure dictated by the product rule. Each term in the resulting expression corresponds to one of the functions being differentiated while the others are kept as they are for that term.

Chain Rule

The chain rule is another fundamental tool in calculus, especially useful when dealing with composite functions. These are functions nested within one another, like \(e^{-2\theta}\) and \(\cos(5\theta)\) in our exercise. The chain rule helps us differentiate such functions by breaking them down into their outer and inner components.

Here's how the chain rule works: If we have a composite function \(y = f(g(x))\), the derivative \(y'\) is determined by multiplying the derivative of the outer function \(f\) (with respect to its argument) by the derivative of the inner function \(g\) (with respect to \(x\)). This gives:

• \(\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\).

In our solution, when differentiating \(e^{-2\theta}\), we think of \(e^u\) where \(u = -2\theta\), and thus, \((e^u)' = e^u\) multiplied by \(u'\). Similarly, \(\cos(5\theta)\) involves a chain of \(\cos(u)\) where \(u = 5\theta\). Getting comfortable with identifying the inner and outer functions makes applying the chain rule seamlessly efficient in even more complex expressions.

Trigonometric Functions

Trigonometric functions, like \(\cos\) and \(\sin\), are ubiquitous in various fields of science and engineering. In calculus, understanding how to differentiate these functions is key.

The basic derivatives for the primary trigonometric functions are:

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

In our problem, we apply this understanding by differentiating \(\cos(5\theta)\). When coupled with the chain rule, we take an additional step:

• The derivative of \(\cos(5\theta)\) becomes \(-\sin(5\theta)\) times the derivative of the inner function, which is \(5\), resulting in \(-5\sin(5\theta)\).

This comprehensive understanding of both the trigonometric identities and the application of the chain rule in trigonometric contexts allows us to handle derivatives of trigonometric expressions confidently and accurately.