Question

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

Short answer

The derivative is \( g'(\theta) = 3(\sin \theta + \cos \theta)^2 (\cos \theta - \sin \theta) \).

Step-by-step solution

Identify the Function Type

The given function is in the form of a composition of functions, specifically the cube of a trigonometric expression: \( g(\theta) = (\sin \theta + \cos \theta)^3 \). This indicates that we need to use the chain rule in conjunction with derivatives of trigonometric functions.

Differentiate the Outer Function

Let \( u = \sin \theta + \cos \theta \). Then, \( g(\theta) = u^3 \). The derivative of \( u^3 \) with respect to \( u \) is \( 3u^2 \). So, the derivative of the outer function is \( 3(\sin \theta + \cos \theta)^2 \).

Differentiate the Inner Function

Differentiate the inner function \( u = \sin \theta + \cos \theta \) with respect to \( \theta \). The derivative is \( \cos \theta - \sin \theta \).

Apply the Chain Rule

Using the chain rule, the derivative \( g'(\theta) \) is given by the product of the derivative of the outer function and the derivative of the inner function: \( g'(\theta) = 3(\sin \theta + \cos \theta)^2 (\cos \theta - \sin \theta) \).

Key concepts

Chain Rule

Calculating derivatives of complex functions often requires breaking down the function into smaller pieces. The chain rule is a fundamental technique in calculus for this purpose. It's useful when dealing with a composition of functions, meaning one function nested inside another.
Imagine you want to differentiate a function like \( g(\theta) = (\sin\theta + \cos\theta)^3 \). Here, the outer function is \( u^3 \) (where \( u = \sin\theta + \cos\theta \)) and the inner function is \( u = \sin\theta + \cos\theta \).
The chain rule tells us to find the derivative of the outer function first with respect to the inner function. Then, multiply this result by the derivative of the inner function with respect to the original variable, \( \theta \).
This step-by-step approach makes it simpler to handle complex derivatives. It's like peeling an onion layer by layer: patiently and methodically.

Trigonometric Functions

Trigonometric functions such as \( \sin \theta \) and \( \cos \theta \) are routine elements in calculus problems. They are especially common in problems involving periodic or wave-like phenomena.
Each trigonometric function has a specific derivative to remember:

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

These derivatives are fundamental since they allow for the calculation of rates of change in trigonometric expressions.
In our exercise, differentiating \( \sin \theta + \cos \theta \) results in \( \cos \theta - \sin \theta \). Each small piece combines to help simplify the larger calculus problem.

Composition of Functions

A composition of functions occurs when one function is applied within another, such as \( (\sin\theta + \cos\theta)^3 \). Here, the expression \( \sin\theta + \cos\theta \) is within the cubic function.
Compositions are tricky; they require methods like the chain rule to differentiate effectively. This structure frequently appears in advanced calculus problems, making it critical to understand.
When dealing with them, remember:

• Identify the inner and outer functions clearly. This clarifies what part to differentiate first.
• Apply the outer derivative to the inner function, then consider the inner function's derivative separately.
• Multiply these derivatives together to find the full derivative of the composition.

This understanding sets a foundation for tackling more intricate functions, ensuring that each component is correctly managed.