Problem 70 - Calculus Single Variable

Question

Calculate the derivative of the given expression. $$ \sin (2 x) $$

Step-by-Step Solution

Answer

The derivative of \( \sin(2x) \) is \( 2\cos(2x) \).

Step 1: Identify the Outer Function

The given expression is \( \sin(2x) \). Here, the outer function is \( \sin(u) \), where \( u = 2x \). We'll need to apply the chain rule to find its derivative.

Step 2: Differentiate the Outer Function

The derivative of \( \sin(u) \) with respect to \( u \) is \( \cos(u) \). So, \( \frac{d}{du}[\sin(u)] = \cos(u) \).

Step 3: Differentiate the Inner Function

The inner function is \( 2x \). The derivative of \( 2x \) with respect to \( x \) is \( 2 \). Therefore, \( \frac{d}{dx}[2x] = 2 \).

Step 4: Apply the Chain Rule

According to the chain rule, the derivative of \( \sin(2x) \) with respect to \( x \) is the derivative of the outer function with respect to the inner function, multiplied by the derivative of the inner function. Thus, \( \frac{d}{dx}[\sin(2x)] = \cos(2x) \cdot 2 \).

Step 5: Simplify the Expression

Multiply the terms to get the final derivative: \( 2\cos(2x) \).

Key Concepts

Chain Rule

The chain rule is an essential tool in calculus for differentiating composite functions, which are functions composed of other functions. If you have a function like \( f(g(x)) \), where \( f \) is the outer function and \( g \) is the inner function, the chain rule states that the derivative of this composite function is the derivative of the outer function with respect to the inner function, multiplied by the derivative of the inner function with respect to \( x \).
To express this mathematically:

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

In the original exercise with \( \sin(2x) \), \( \sin \) is the outer function and \( 2x \) is the inner. The chain rule helps in differentiating \( \sin(2x) \) to get \( 2\cos(2x) \). By decomposing the function into its parts, it becomes clear how each contributes to the derivative result.

Trigonometric Differentiation

Trigonometric functions are fundamental in calculus and have specific differentiation rules that simplify the process.
For instance, the derivative of \( \sin(x) \) is \( \cos(x) \), while the derivative of \( \cos(x) \) is \( -\sin(x) \).
These basic derivatives are crucial for problems involving trigonometric expressions. In applying these rules to the problem of \( \sin(2x) \), recognizing that we start with the trigonometric identity allows us to focus on applying the chain rule effectively.
When solving more complex expressions, it's important to remember:

• Each trigonometric function has a specific derivative rule.
• Pay attention to the composition of functions, as seen in \( \sin(2x) \) where \( 2x \) is the argument.
• Use identities where necessary, for simplifying and differentiating trigonometric expressions.

Having a strong grasp on these derivatives allows one to confidently tackle more advanced calculus problems.

Multivariable Calculus

Multivariable calculus extends differentiation and integration concepts to functions of several variables. This field is essential for areas that require analyzing how several variables relate interdependently.
When dealing with derivatives in a multivariable setting, it's possible to encounter situations similar to the current problem but expanded to include more dimensions.

• Partial derivatives become crucial, as they involve differentiating with respect to one variable while keeping others constant.
• Real-world scenarios often demand understanding of how changing one variable affects a collective system.

If \( F(x, y) = \sin(2x) + y \), finding the partial derivative with respect to \( x \), we still apply similar steps using the chain rule.
The concept of directional derivatives also comes into play, providing rates of change in specific directions. Understanding single-variable calculus concepts like the chain rule assists greatly when advancing into the realms of multivariable calculus, emphasizing how foundational understanding builds towards complexity.