Question

Compute the derivative of the given function. $$f(\theta)=\theta^{3} \sin \theta+\frac{\sin \theta}{\theta^{3}}$$

Short answer

The derivative is \( f'(\theta) = 3\theta^2 \sin \theta - 3\theta^{-4} \sin \theta + \theta^3 \cos \theta + \theta^{-3} \cos \theta \).

Step-by-step solution

Identify the function parts

The given function is \( f(\theta) = \theta^3 \sin \theta + \frac{\sin \theta}{\theta^3} \). It can be split into two terms: the first term is \( \theta^3 \sin \theta \), and the second term is \( \frac{\sin \theta}{\theta^3} \). Each term will be differentiated separately.

Differentiate the first term

For the term \( \theta^3 \sin \theta \), apply the product rule. If \( u = \theta^3 \) and \( v = \sin \theta \), then the derivative is \( u'v + uv' \). Compute:- \( u' = 3\theta^2 \)- \( v' = \cos \theta \)Therefore, the derivative of \( \theta^3 \sin \theta \) is \( 3\theta^2 \sin \theta + \theta^3 \cos \theta \).

Differentiate the second term

For the term \( \frac{\sin \theta}{\theta^3} \), rewrite it as \( \theta^{-3} \sin \theta \). Again, apply the product rule, where:- \( u = \theta^{-3} \) and \( v = \sin \theta \).- \( u' = -3\theta^{-4} \) and \( v' = \cos \theta \).The derivative of \( \theta^{-3} \sin \theta \) is \( -3\theta^{-4} \sin \theta + \theta^{-3} \cos \theta \).

Combine the derivatives

Combine the derivatives from Step 2 and Step 3 to find the derivative of the whole function \( f(\theta) \). This yields:\[ f'(\theta) = (3\theta^2 \sin \theta + \theta^3 \cos \theta) + (-3\theta^{-4} \sin \theta + \theta^{-3} \cos \theta) \].

Simplify the expression

Combine like terms in the expression:\[ f'(\theta) = 3\theta^2 \sin \theta - 3\theta^{-4} \sin \theta + \theta^3 \cos \theta + \theta^{-3} \cos \theta \].Notice that no further simplification is possible, so this is the simplified form of the derivative.

Key concepts

Product Rule

When dealing with derivatives, especially involving functions with multiple parts, the product rule is an essential tool. The product rule states that the derivative of a product of two functions is given by:

• For functions \( u \) and \( v \), the derivative \( (uv)' \) is \( u'v + uv' \).

In simple terms, differentiate the first function, leave the second function as it is, then add the term where the first function is as it is, and differentiate the second function.

Let's see an example. Take the function: \( \theta^3 \sin \theta \).

- Here, \( u = \theta^3 \) and \( v = \sin \theta \).- Differentiate \( u \) to get \( u' = 3\theta^2 \).- Differentiate \( v \) to obtain \( v' = \cos \theta \).Using the product rule, we find the derivative of this term as:

• \( 3\theta^2 \sin \theta + \theta^3 \cos \theta \).

Trigonometric Functions

Trigonometric functions like sine and cosine are not only central to trigonometry but also play an important role in calculus. It's essential to understand how to differentiate these functions, as they frequently appear in various problems.

Recall these basic derivatives:

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

In the exercise, trigonometric functions appear in both terms of our function. For the product \( \theta^3 \sin \theta \), the sine function is part of our effort in applying the product rule, leading to the derivative \( \cos \theta \).

Similarly, in the term \( \frac{\sin \theta}{\theta^3} \), rewriting it in the form of \( \theta^{-3} \sin \theta \) shows us the sine function again, where we also calculate \( \cos \theta \) as part of the derivative.

Remembering these fundamental derivatives makes tackling problems involving trigonometric functions a smoother process.

Chain Rule

The chain rule is used for differentiating composite functions. Although not explicitly highlighted in every step of this specific exercise, it's worth knowing because often the expressions we work with might appear as compositions.

The rule can be stated as: if a function \( y = f(g(x)) \), then its derivative is given by:

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

Consider the scenario of a function being inside another, like \( \sin(\theta^3) \); though not present here, it serves as a fine example. Here, the chain rule helps by first finding the derivative of the outer function, \( \sin(x) \), and multiplying it by the derivative of the inner function \( \theta^3 \).

Being comfortable with the chain rule enhances your ability to recognize and solve more complex problems, especially those involving nested functions or where transformations occur. It is a crucial skill in calculus that complements both the product rule and a good grasp of trigonometric functions.