Question

Find the indicated derivative. $$ \frac{d y}{d x}, \text { where } y=\left(\frac{\sin x}{\cos 2 x}\right)^{3} $$

Short answer

The derivative is obtained using the chain and quotient rules, with simplifications involving trigonometric identities.

Step-by-step solution

Recognize the Chain Rule

To find the derivative of the function, we first notice that it is a composition of functions. The expression \( y = \left(\frac{\sin x}{\cos 2x}\right)^{3} \) indicates the use of the chain rule because we have an outer function \( u^{3} \) and an inner function \( u = \frac{\sin x}{\cos 2x} \). Our task is to differentiate the composition.

Apply the Chain Rule

Using the chain rule, the derivative of \( y \) with respect to \( x \) is \( \frac{dy}{dx} = 3 \left(\frac{\sin x}{\cos 2x}\right)^{2} \cdot \frac{d}{dx} \left(\frac{\sin x}{\cos 2x}\right) \). This means we must now find the derivative of the inner function \( \frac{\sin x}{\cos 2x} \).

Differentiate the Inner Function Using the Quotient Rule

The inner function \( \frac{\sin x}{\cos 2x} \) requires the quotient rule for differentiation. The quotient rule formula is \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\cdot u' - u \cdot v'}{v^2} \). Here, \( u = \sin x \) and \( v = \cos 2x \). Find derivatives: \( u' = \cos x \) and \( v' = -2\sin 2x \).

Compute Derivative of Inner Function

Substitute identified functions and their derivatives into the quotient rule: \[ \frac{d}{dx}\left(\frac{\sin x}{\cos 2x}\right) = \frac{\cos 2x \cdot \cos x - \sin x \cdot (-2\sin 2x)}{\cos^2 2x} \]. Simplify using trigonometric identities: \( \sin 2x = 2\sin x \cos x \), thus \( 2\sin x \sin 2x = 4\sin x^2 \cos x \). Simplify the fraction.

Substitute Inner Derivative Back into Chain Rule Expression

Substitute the result from Step 4 into the chain rule expression from Step 2, giving us: \[ \frac{dy}{dx} = 3 \left(\frac{\sin x}{\cos 2x}\right)^{2} \cdot \frac{\cos 2x \cdot \cos x + 2\sin x \cdot \sin 2x}{\cos^2 2x} \]. Simplify this expression by cancelling or reducing terms where possible.

Key concepts

Chain Rule

The chain rule is a fundamental theorem in calculus used to differentiate compositions of functions. When you have a function nested inside another, such as a power of a fraction, the chain rule helps you differentiate efficiently.
In our exercise, we have an outer function, specifically the cubic function, and an inner function, which is a trigonometric fraction.
The formula for the chain rule is:

• If you have a composite function of the form \( y = f(g(x)) \), then the derivative \( \frac{dy}{dx} \) is \( f'(g(x)) \cdot g'(x) \).

Here, the outer function \( u^3 \) leads to the derivative \( 3u^2 \) when applying the chain rule. The inner function requires separate differentiation.
Recognizing this interplay between functions is key to applying the chain rule correctly.

Quotient Rule

The quotient rule is vital when differentiating ratios, usually involving trigonometric functions or polynomial fractions. When you have a numerator and denominator both as functions of \( x \), the rule aids in finding the derivative efficiently.
In mathematical terms, the quotient rule states:

• If you have a function \( h(x) = \frac{u(x)}{v(x)} \), then the derivative \( \frac{d}{dx}h(x) \) is \( \frac{v(x)u'(x) - u(x)v'(x)}{v(x)^2} \).

Here, the inner function \( \frac{\sin x}{\cos 2x} \) makes use of the quotient rule.
You differentiate the top part \( \sin x \) to get \( \cos x \), and differentiate the bottom \( \cos 2x \) to find \(-2\sin 2x \).
Substitute these values back into the quotient rule formula to find the derivative of the inner function.

Trigonometric Identities

Trigonometric identities are extremely helpful when simplifying derivatives, especially those involving sine and cosine functions.
After applying the quotient rule to our exercise, they assist in reducing the complexity of the expression.
Two key identities often used include:

• \( \sin 2x = 2\sin x \cos x \)
• Power-reduction identities that help refocus trigonometric expressions into simpler forms

In our exercise, using \( \sin 2x = 2\sin x \cos x \) helped simplify the numerator of the derived function.
This step is crucial to making complex derivatives manageable, as it aids in combining like terms and reducing any potential terms, ensuring an accurate solution.