Problem 68 - Calculus Single Variable

Question

Calculate the derivative of the given expression. $$ \sin ^{3}(x) $$

Step-by-Step Solution

Answer

The derivative of \( \sin^{3}(x) \) is \( 3\sin^{2}(x)\cos(x) \).

Step 1: Understand the Expression

The given expression is \( \sin^{3}(x) \), which means \( (\sin(x))^3 \). This is a composition of functions: an outer function \( u^3 \) and an inner function \( \sin(x) \).

Step 2: Apply the Chain Rule

To find the derivative of a composite function \( f(g(x)) \), use the chain rule: \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \). Here, we let \( u = \sin(x) \) and then our expression becomes \( u^3 \).

Step 3: Differentiate the Outer Function

The outer function is \( u^3 \). The derivative of \( u^3 \) with respect to \( u \) is \( 3u^2 \).

Step 4: Differentiate the Inner Function

The inner function is \( \sin(x) \). The derivative of \( \sin(x) \) with respect to \( x \) is \( \cos(x) \).

Step 5: Combine Derivatives Using the Chain Rule

According to the chain rule, the derivative of \( \sin^{3}(x) \) is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function: \[ \frac{d}{dx} (\sin^{3}(x)) = 3(\sin(x))^2 \cdot \cos(x) \].

Step 6: Simplify the Expression

The derivative simplifies to: \[ 3\sin^{2}(x)\cos(x) \]. This is the final simplified expression for the derivative.

Key Concepts

Chain Rule

The chain rule is a crucial concept in calculus, allowing us to find the derivative of composite functions, which are functions defined by the composition of two or more other functions. When applied, it helps in differentiating expressions where one function is nested inside another. In mathematical terms, if you have a function expressed as \( f(g(x)) \), the chain rule states:

• \( (f(g(x)))' = f'(g(x)) \times g'(x) \)

This rule essentially tells us to take the derivative of the outer function and multiply it by the derivative of the inner function.
In our specific problem, \( 0sin^{3}(x) \), the outer function is \(u^3\) and the inner function is \(sin(x)\). It's like peeling an onion; we first differentiate the outer layer, then work our way inward by using the chain rule to connect the parts. This results in multiplying the derivative of the outer function, evaluated at the inner function, by the derivative of the inner function.

Derivative of Sine Function

The derivative of the sine function is a foundational derivative in trigonometry and calculus. For the sine function, \( 0sin(x) \), its derivative is well-known:

• The derivative \( 0\frac{d}{dx} ( 0sin(x)) = cos(x) \).

This result means that as the input \(x\) changes, the rate of change of \( 0sin(x) \) is represented by \( 0cos(x) \).
In our exercise, the derivative of \( 0sin(x) \) plays a pivotal role as it is part of the inner function we worked with. When using the chain rule, this derivative becomes crucial and multiplies with the derivative of the composed outer function to give the full derivative of \( 0sin^{3}(x) \). Thus, ensuring you understand this simple derivative helps break down more complex problems involving trigonometric functions.

Composite Functions

Composite functions are those which are formed by the composition of two or more other functions. It essentially means you are applying one function to the results of another function.
For instance, in the expression \( 0sin^{3}(x) \), this is really \(( 0sin(x))^3\). Recognizing this form allows us to break down the problem into simpler parts.
In general:

• The expression \( f(g(x)) \) is an example of a composite function.
• Here, \( f(u) \) might represent an outer function like \( u^3 \), while \( g(x) \) is the inner function \( 0sin(x) \).

Understanding composite functions is key in calculus because many real-world applications involve nested functions. The ability to apply the chain rule to these composites allows for easily handling derivatives in complex problems, giving insight into how variables interact within functional systems.