Question

Use the identity \(\sin 2 x=2 \sin x \cos x\) to find \(\frac{d}{d x}(\sin 2 x) .\) Then use the identity \(\cos 2 x=\cos ^{2} x-\sin ^{2} x\) to express the derivative of \(\sin 2 x\) in terms of \(\cos 2 x\).

Short answer

In conclusion, the derivative of \(\sin 2x\) is \(\frac{d}{dx}(\sin 2x) = 2\cos 2x\), expressed in terms of \(\cos 2x\).

Step-by-step solution

Find the derivative of \(\sin 2x\) using identity 1

Recall that the derivative of \(\sin x\) is \(\cos x\), and the derivative of \(\cos x\) is \(-\sin x\). Using identity 1, we rewrite the expression \(\sin 2x = 2\sin x\cos x\). Let's now find the derivative of this expression with respect to \(x\): \(\frac{d}{dx}(\sin 2x) = \frac{d}{dx}(2\sin x\cos x)\) Now, use the product rule for differentiation, which states that if you have a function \(u(x)v(x)\), then the derivative is \(\frac{d}{dx}(u(x)v(x)) = u'(x)v(x) + u(x)v'(x)\). In our case, \(u(x) = 2\sin x\) and \(v(x) = \cos x\). So the derivatives are \(u'(x) = 2\cos x\) and \(v'(x) = -\sin x\). Apply the product rule: \(\frac{d}{dx}(\sin 2x) = (2\cos x)(\cos x) + (2\sin x)(-\sin x)\)

Rewrite the derivative using identity 2

Now, we will use identity 2 \(\cos 2x = \cos^2 x - \sin^2 x\) to rewrite the derivative of \(\sin 2x\) in terms of \(\cos 2x\). We can write the derivative as: \(\frac{d}{dx}(\sin 2x) = 2(\cos^2 x - \sin^2 x)\) Now, using identity 2, replace the expression inside the parenthesis: \(\frac{d}{dx}(\sin 2x) = 2(\cos 2x)\) That's the result. The derivative of \(\sin 2x\), expressed in terms of \(\cos 2x\), is \(2\cos 2x\).

Key concepts

Trigonometric Identities

Trigonometric identities are vital tools in calculus. They are equations involving trigonometric functions that are true for every value of the variables involved. These identities help simplify expressions and solve equations. In the context of the derivative of trigonometric functions, understanding identities such as

• the double-angle identity: \(\sin 2x = 2\sin x \cos x\) and \(\cos 2x = \cos^2 x - \sin^2 x\)

is crucial.The double-angle identity \(\sin 2x = 2\sin x \cos x\) allows us to express \(\sin 2x\) as a product of \(\sin x\) and \(\cos x\). Similarly, \(\cos 2x = \cos^2 x - \sin^2 x\) helps in rewriting expressions involving \(\cos 2x\). By substituting these identities, we can transform complex trigonometric expressions into simpler forms. This is particularly helpful when we compute derivatives of trigonometric functions, as it can simplify the differentiation process by allowing the application of the basic differentiation rules.

Product Rule in Calculus

The product rule is essential when differentiating expressions that are the product of two functions. If you have a function defined as \(y = u(x)\cdot v(x)\), the product rule states that its derivative \(\frac{d}{dx}(y)\) is:

• \(u'(x)v(x) + u(x)v'(x)\)

This means you take the derivative of the first function \(u(x)\) while keeping the second \(v(x)\) constant, and add it to the derivative of the second function \(v(x)\) while keeping the first \(u(x)\) constant.

For example, using the product rule for \(\sin 2x = 2\sin x \cos x\), where \(u(x) = 2\sin x\) and \(v(x) = \cos x\):

• First, find the derivatives: \(u'(x) = 2\cos x\) and \(v'(x) = -\sin x\).
• Then, apply the product rule: \[\frac{d}{dx}(\sin 2x) = (2\cos x)(\cos x) + (2\sin x)(-\sin x)\]

This technique is widely used in calculus to easily differentiate products of functions, yielding the correct derivative through a careful application of simple rules.

Chain Rule in Calculus

The chain rule is an extremely useful formula for finding the derivative of composite functions. It is especially helpful when one function is nested inside another. The chain rule states:

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

This allows you to differentiate complex expressions by breaking them down into simpler parts. Essentially, you're finding the derivative of the outer function \(f\) evaluated at the inner function \(g\), and multiplying it by the derivative of the inner function.

While the original exercise focused on the application of trigonometric identities and the product rule, the chain rule often complements these methods. For example, when dealing with expressions like \(\sin(2x)\), you could view \(2x\) as the inner function and apply the chain rule to differentiate directly:

• The derivative of \(\sin(2x)\) using the chain rule is \(\cos(2x)\cdot 2\).

Thus, employing the chain rule simplifies many differentiation problems, particularly when dealing directly with functions like \(\sin(kx)\) or more complex nested functions.