Problem 35 - Thomas Calculus

Question

Find the derivatives of the functions in Exercises \(19-38\) $$ r=\sin \left(\theta^{2}\right) \cos (2 \theta) $$

Step-by-Step Solution

Answer

The derivative is \(r' = 2\theta \cos(\theta^2) \cos(2\theta) - 2\sin(\theta^2) \sin(2\theta)\).

Step 1: Identify the Rule of Differentiation

The function is a product of two functions: \(r = \sin(\theta^2)\) and \(\cos(2\theta)\). To differentiate this function, we need to use the product rule of differentiation.

Step 2: Apply the Product Rule

The product rule states that if \(u\) and \(v\) are functions of \(\theta\), then the derivative of the product \(uv\) is given by \((uv)' = u'v + uv'\). Let \(u = \sin(\theta^2)\) and \(v = \cos(2\theta)\). We need to find \(u'\), \(v\), and \(v'\), \(u\).

Step 3: Differentiate \(u = \sin(\theta^2)\)

To differentiate \(u = \sin(\theta^2)\) with respect to \(\theta\), use the chain rule:\[ u' = \cos(\theta^2) \cdot (\theta^2)' = \cos(\theta^2) \cdot 2\theta = 2\theta \cos(\theta^2) \].

Step 4: Differentiate \(v = \cos(2\theta)\)

Differentiate \(v = \cos(2\theta)\) with respect to \(\theta\) using the chain rule:\[ v' = -\sin(2\theta) \cdot (2\theta)' = -\sin(2\theta) \cdot 2 = -2\sin(2\theta) \].

Step 5: Substitute and Simplify

Substitute the derivatives \(u'\) and \(v'\) back into the product rule formula:\[ r' = (2\theta \cos(\theta^2)) \cos(2\theta) + \sin(\theta^2)(-2\sin(2\theta)) \].Simplify the expression:\[ r' = 2\theta \cos(\theta^2) \cos(2\theta) - 2\sin(\theta^2) \sin(2\theta) \].

Key Concepts

Chain Rule

In calculus, the chain rule is an essential tool for finding the derivatives of composite functions. It allows us to differentiate a function that is nested inside another function, which is common in more complicated expressions. When you have a function of the form \( g(f(x)) \), the chain rule states that the derivative is the product of the derivative of the outer function, evaluated at the inner function, and the derivative of the inner function.

• In formula terms, it can be written as \( (g(f(x)))' = g'(f(x)) \cdot f'(x) \).
• The chain rule is particularly useful when dealing with functions like \( \sin(\theta^2) \) or \( \cos(2\theta) \), where an operation like squaring or multiplying by a constant is applied to a variable before applying the sine or cosine functions.

By understanding and applying the chain rule, you can effectively tackle derivatives of more complex functions. It's a stepping stone to mastering calculus exercises, especially those involving function compositions.

Trigonometric Functions Differentiation

Differentiating trigonometric functions is a common task in calculus, and knowing the derivatives of basic trigonometric functions is crucial.

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

When combined with other operations like multiplication or function composition, these derivatives form the basis for solving more complex calculus problems.
For example, in the function \( \sin(\theta^2) \), you apply the chain rule to differentiate it, as it involves \( \theta^2 \) being placed inside the sine function. Similarly, for \( \cos(2\theta) \), the constant factor 2 requires the chain rule for differentiation, leading to the derivative involving \( -2\sin(2\theta) \).
Grasping these concepts offers a solid understanding to tackle problems involving trigonometric functions in various calculus exercises.

Calculus Exercises

Engaging with calculus exercises is key to mastering differentiation and integration techniques. Calculus exercises often involve multiple steps and the application of various rules such as the chain rule or product rule.

• Start by clearly identifying the type of functions involved. Are they simple polynomials, trigonometric, or a combination?
• Determine which differentiation rules apply. For example, use the product rule when differentiating products like \( \sin(\theta^2) \cdot \cos(2\theta) \).
• Carefully apply these rules, making sure each step logically follows the previous.

Simplifying expressions after differentiating is crucial in obtaining the clean and correct final answer. Exercises like the one provided help build an intuitive understanding of calculus, making future problems seem less daunting. Regular practice with varied exercises ensures that concepts such as trigonometric differentiation and chain rule application become second nature.