Question

$$ \begin{array}{l} \text { . Use the trigonometric identity } \sin 2 x=2 \sin x \cos x\\\ \text { along with the Product Rule to find } D_{x} \sin 2 x \text { . } \end{array} $$

Short answer

The derivative is \( 2(\cos^2 x - \sin^2 x) \).

Step-by-step solution

Write Down the Derivative to Find

We need to find the derivative of \( \sin 2x \) with respect to \( x \). This is represented as \( D_{x}(\sin 2x) \).

Use Trigonometric Identity

Recall the identity \( \sin 2x = 2 \sin x \cos x \). This identity will help us express \( \sin 2x \) in terms of \( \sin x \) and \( \cos x \), which is useful for applying the product rule.

Apply the Product Rule

The product rule states that if \( u(x) \) and \( v(x) \) are functions, then the derivative \( D_x(uv) = u'v + uv' \). For \( \sin 2x = 2 \sin x \cos x \), let \( u = 2 \sin x \) and \( v = \cos x \).

Differentiate Each Function

Calculate the derivatives \( u' \) and \( v' \): - \( u = 2 \sin x \), so \( u' = 2 \cos x \)- \( v = \cos x \), so \( v' = -\sin x \)

Apply Values to Product Rule Formula

Using the product rule formula: \( D_x(2 \sin x \cos x) = (2 \cos x)(\cos x) + (2 \sin x)(-\sin x) \)

Simplify the Expression

Simplify the expression from Step 5:\( = 2(\cos^2 x) - 2(\sin^2 x) \)\( = 2(\cos^2 x - \sin^2 x) \)

Write the Final Answer

Thus, \( D_{x}(\sin 2x) = 2(\cos^2 x - \sin^2 x) \).

Key concepts

Trigonometric Identities

Trigonometric identities are formulas that involve trigonometric functions and are true for all values of the variables for which the functions are defined. One of the most useful identities is the double angle identity for sine, given by \( \sin 2x = 2 \sin x \cos x \). This identity allows us to express the sine of a double angle in terms of sine and cosine of the original angle.
In our exercise, we use this identity to rewrite \( \sin 2x \) as a product of \( \sin x \) and \( \cos x \). This makes it easier to apply calculus concepts like the product rule, which is often necessary when dealing with expressions involving multiple functions.

Product Rule

The product rule is a fundamental rule of differentiation in calculus. It helps when you need to take derivatives of expressions that are products of two functions. The product rule states: if \( u(x) \) and \( v(x) \) are differentiable functions, then the derivative of their product is given by \( D_x(uv) = u'v + uv' \).
In our example, after using the identity \( \sin 2x = 2 \sin x \cos x \), we identified the functions as \( u = 2 \sin x \) and \( v = \cos x \). Each of these functions can be differentiated separately. By applying the product rule, we can then find the overall derivative of the expression \( 2 \sin x \cos x \).
This step-by-step approach ensures precision and helps avoid common mistakes in differentiation.

Derivatives

Derivatives represent the rate of change of a function with respect to a variable. It is one of the core concepts of calculus and provides critical information about how a function behaves.
In the context of our exercise, finding the derivative \( D_x(\sin 2x) \) involves using both trigonometric identities and the product rule. First, we redefine \( \sin 2x \) in terms of \( \sin x \) and \( \cos x \), so that we can easily differentiate each part.
The derivative of \( u = 2 \sin x \) is \( u' = 2 \cos x \), and the derivative of \( v = \cos x \) is \( v' = -\sin x \). These individual derivatives form the building blocks in applying the product rule and thus in finding the complete derivative.

Calculus

Calculus is a branch of mathematics that deals with continuous change, and it is fundamentally built around concepts like derivatives and integrals. The power of calculus lies in its ability to model and solve problems that involve dynamic quantities.
In this problem, calculus allows us to calculate the rate at which the function \( \sin 2x \) changes with respect to \( x \). We use differentiating rules, like the product rule, and identities, such as the trigonometric double angle identity, to find this rate efficiently.
By understanding the foundational concepts of calculus, we can apply these principles to complex expressions. This understanding is crucial because it gives us the flexibility to manipulate and find solutions for a wide variety of mathematical and real-world problems.