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Chapter 2: Problem 39

Find the derivative of the transcendental function. $$ y=\frac{3(1-\sin x)}{2 \cos x} $$

Short Answer

Expert verified
The derivative of the function \(y=\frac{3(1-\sin x)}{2 \cos x}\) is \(y'=\frac{3(2-\sin x)}{2 \cos^2 x}\).

Step by step solution

01

Quotient Rule Preparation

Differentiate the numerator and the denominator separately. The derivative of \(3(1-\sin x)\) using the chain rule results in -3\(\cos x\). The derivative of \(2 \cos x\) using the basic derivative rule results in -2\(\sin x\).
02

Apply Quotient Rule

According to the quotient rule, \((f/g)'\) is equal to \((g f'-f g') / g^2\). Applying this rule to our function, we have \(y' = \frac{-2\sin x \cdot -3\cos x - -2\sin x \cdot 3(1-\sin x)}{(2\cos x)^2}\). Simplify this expression by multiplying and combining like terms.
03

Simplify the Expression

The simplified version of the derivative is \(\frac{6\sin^2 x + 6 \cos^2 x - 6\sin x}{4\cos^2 x}\). Use the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\) to further simplify.
04

Further Simplification Using Trigonometric Identity

Applying the identity results in \(\frac{6 - 6\sin x}{4\cos^2 x}\) or \( \frac{3(2-\sin x)}{2 \cos^2 x}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quotient Rule
When you have a function that is the quotient, or division, of two other functions, the quotient rule is your go-to tool for finding the derivative. This rule states that if you have a function defined as \(y = \frac{f(x)}{g(x)}\), its derivative is given by:
  • \(y' = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2} \)
Here, \(f(x)\) is the numerator part and \(g(x)\) is the denominator part of your function.

In our exercise, the function is \(y=\frac{3(1-\sin x)}{2 \cos x}\). To apply the quotient rule:
  • Differentiate the numerator, \(3(1-\sin x)\), giving us \(-3\cos x\).
  • Differentiate the denominator, \(2 \cos x\), giving us \(-2\sin x\).
    • Insert these values into the quotient rule formula to find the derivative.
    Thus, the derivative involves taking these differentiated parts and plugging them back into the quotient rule formula to find the correct expression.
Chain Rule
In calculus, the chain rule is vital for finding derivatives, especially when dealing with composite functions. Composite functions are those where one function is nested within another, expressed as \(f(g(x))\). The chain rule tells us how to differentiate these nested functions:
  • If \(y = f(g(x))\), then the derivative of \(y\) with respect to \(x\) is \(y' = f'(g(x)) \, g'(x)\).
Within our exercise, the chain rule was crucial in finding the derivative of the numerator \(3(1-\sin x)\).

Breaking it down:
  • The outer function is \(3u\), where \(u = 1 - \sin x\).
  • The derivative of \(3u\) with respect to \(u\) is \(3\).
  • For \(1-\sin x\), the derivative is \(-\cos x\).
Therefore, applying the chain rule yields the derivative of the numerator as \(-3\cos x\).
Understanding the chain rule aids tremendously in tackling more complex derivatives.
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that hold true for all values involved. They are incredibly useful for simplifying derivatives and other expressions.

A key trigonometric identity used here is the Pythagorean identity:
  • \(\sin^2 x + \cos^2 x = 1\)
In the context of our solution, after applying the quotient rule, the expression obtained was \(\frac{6\sin^2 x + 6 \cos^2 x - 6\sin x}{4\cos^2 x}\).
By substituting \(\sin^2 x + \cos^2 x = 1\), we can transform the expression:
  • Replace \(\sin^2 x + \cos^2 x\) with \(1\), simplifying to \(\frac{6 - 6\sin x}{4\cos^2 x}\).
Trigonometric identities are crucial for further simplifying trigonometric derivatives and ensuring we obtain the simplest possible expression.

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