Question

Use the identity \(\cos 2 x=\cos x \cos x-\sin x \sin x\) to find the derivative of \(\cos 2 x,\) Express the derivative in terms of \(\sin 2 x\)

Short answer

The derivative of \( \cos 2x \), expressed in terms of \( \sin 2x \), is \( -2\sin 2x \).

Step-by-step solution

Rewrite the Function Using Trigonometric Identity

We know that \( \cos 2x = \cos^2x - \sin^2x \). Our first step will be to rewrite the function in this form.

Take the derivative

Next, we need to differentiate the function. Using the chain rule, the derivative of \( \cos u \) where \( u \) is a function of \( x \), is \( -u' \sin u \) . Applying this to our function we get, \(y' = -2\sin2x \). Considering sine function, we get \( y' = -2\sin(x)\cos(x) - 2\cos(x)\sin(x) = -2\sin 2x \)

Express the derivative in terms of \( \sin 2x \)

We have already obtained the derivative in the form of \( -2\sin 2x \), which is the required final form.

Key concepts

Understanding Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the involved variables. These identities are the backbone of trigonometry and are pivotal in solving complex problems in calculus. One such identity is the double angle formula for cosine, which expresses \(\cos 2x\) as \(\cos^2x - \sin^2x\). This identity allows us to rewrite trigonometric expressions in different forms, which can simplify the process of taking derivatives or evaluating integrals.

In the context of derivatives, when faced with a function like \(\cos 2x\), recognizing it as \(\cos^2x - \sin^2x\) instead of a single trigonometric function is crucial. By leveraging this identity, differentiation becomes a straightforward task because you are dealing with basic functions whose derivatives are well known. Trigonometric identities can also facilitate simplification when the expressions are complex, allowing students to break down the problem into more manageable pieces. Understanding and memorizing these identities will significantly aid in solving calculus problems.

Applying the Chain Rule in Calculus

The chain rule is a fundamental theorem in calculus that is used to find the derivative of a composite function. In simple terms, if a variable \(u\) depends on \(x\) and another function \(y\) depends on \(u\), then the rate of change of \(y\) with respect to \(x\) can be found by multiplying the rate of change of \(y\) with respect to \(u\) by the rate of change of \(u\) with respect to \(x\). Symbolically, the chain rule is expressed as \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\).

When taking the derivative of \(\cos 2x\) using the chain rule, we identify \(u = 2x\) and \(y = \cos u\). By taking the derivative of \(y\) with respect to \(u\) and then multiplying it by the derivative of \(u\) with respect to \(x\), we get our solution. It's important to remember that the chain rule provides a methodical way to tackle the differentiation of functions that are compositions of other functions, simplifying the process and ensuring correctness in the calculations.

Differentiation of Trigonometric Functions

Differentiating trigonometric functions is an essential skill in calculus. Each trigonometric function has a specific derivative that students must learn. For example, the derivative of \(\sin x\) is \(\cos x\), and the derivative of \(\cos x\) is \(\-\sin x\). These derivatives are derived from the limit definitions and have proofs rooted in the unit circle.

In case of the exercise at hand, once the function \(\cos 2x\) is written in terms of \(\cos^2x - \sin^2x\), we apply the differentiation rules for basic functions. Since the derivative of \(\sin x\) is \(\cos x\) and that of \(\cos x\) is \(\-\sin x\), when we apply the derivative to \(\cos^2x - \sin^2x\), we have to use the chain rule to account for the inner function \(2x\). This results in the derivative being in terms of \(\sin 2x\). Recognizing these patterns and rules helps students simplify and solve derivatives efficiently. Emphasizing the understanding of these basic derivatives makes the process of solving more complex derivatives much less intimidating.