Chapter 2: Problem 27
Find the indicated derivative. \(\frac{d y}{d x},\) where \(y=\left(\frac{\sin x}{\cos 2 x}\right)^{3}\)
Short Answer
Expert verified
Derive using Chain Rule and Quotient Rule: \( \frac{dy}{dx} = 3 \left( \frac{\sin x}{\cos 2x} \right)^2 \cdot \frac{\cos x}{\cos^2 2x} \).
Step by step solution
01
Apply the Chain Rule
Since we need to derive the function \[ y = igg(\frac{\sin x}{\cos 2x}\bigg)^3 \]we first apply the Chain Rule for derivatives. The Chain Rule states that:\[ \frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x) \]Here, let \( u = \frac{\sin x}{\cos 2x} \) and consider \( y = u^3 \). Then,\[ \frac{dy}{du} = 3u^{2} \]. We will calculate \( \frac{du}{dx} \) next.
02
Differentiate the Inner Function
Now find the derivative of the inner function\[ u = \frac{\sin x}{\cos 2x} \]using the Quotient Rule. The Quotient Rule states that:\[ \frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \]For \( f(x) = \sin x \), we have \( f'(x) = \cos x \). For \( g(x) = \cos 2x \), use the Chain Rule: \( g'(x) = -2\sin 2x \). Therefore,\[ \frac{du}{dx} = \frac{ \cos x \cdot \cos 2x + 2 \sin x \cdot \sin 2x }{\cos^2 2x} \]. Use the trigonometric identity: \( \cos(A-B) = \cos A \cos B + \sin A \sin B \). Thus,\[ \frac{du}{dx} = \frac{\cos(x-2x)}{\cos^2 2x} = \frac{\cos(-x)}{\cos^2 2x} = \frac{\cos x}{\cos^2 2x} \].
03
Combine Results to Get the Derivative
Using the results from the previous steps, substitute back into the Chain Rule:\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 3 \left( \frac{\sin x}{\cos 2x} \right)^2 \cdot \frac{\cos x}{\cos^2 2x} \].Simplify as needed:\[ \frac{dy}{dx} = 3 \left( \frac{\sin x}{\cos 2x} \right)^2 \cdot \frac{\cos x}{\cos^2 2x} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quotient Rule
When dealing with derivatives of functions that are expressed as a ratio of two different functions, the Quotient Rule becomes an essential tool. It provides a systematic approach for differentiating such functions. In mathematical terms, if we have two functions, \( f(x) \) and \( g(x) \), where \( g(x) eq 0 \), the derivative of their ratio is given by:
- \( \frac{d}{dx}\left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \)
Trigonometric Identities
Trigonometric identities simplify the calculus related to trigonometric functions. In this exercise, we utilized one of the fundamental identities: the cosine angle difference identity:
- \( \cos(A-B) = \cos A \cos B + \sin A \sin B \)
Differentiation Techniques
There are multiple differentiation techniques, and choosing the right one depends on the form of the function you are working with. For composite functions, like \( y = (\frac{\sin x}{\cos 2x})^3 \), the Chain Rule is invaluable. Here's how it works:
- The Chain Rule states: \( \frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x) \).
- In our case, \( u = \frac{\sin x}{\cos 2x} \) and \( y = u^3 \).