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In Exercises \(1-18,\) find \(d y / d x\) $$ g(x)=\frac{\cos x}{\sin ^{2} x} $$

Short Answer

Expert verified
The derivative is \( \frac{-(1 + \cos^2 x)}{\sin^3 x} \).

Step by step solution

01

Understand the Function

The function given is a ratio of two trigonometric functions, where the numerator is \( \cos x \) and the denominator is \( \sin^2 x \). Essentially, this function can be rewriten as \( g(x) = \cos x \cdot (\sin x)^{-2} \).
02

Apply the Quotient Rule

Since the function is a quotient of two functions, we use the quotient rule, which states that if \( u(x) = \cos x \) and \( v(x) = \sin^2 x \), then \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2} \).
03

Differentiate the Numerator

Differentiate the numerator: \( u(x) = \cos x \) gives \( u'(x) = -\sin x \).
04

Differentiate the Denominator

Differentiate the denominator: \( v(x) = \sin^2 x \) requires the chain rule. Hence, \( v'(x) = 2\sin x \cdot \cos x \).
05

Apply the Quotient Rule Formula

Substitute the derivatives into the quotient rule formula: \[ \frac{d}{dx}\left(\frac{\cos x}{\sin^2 x}\right) = \frac{\sin^2 x \cdot (-\sin x) - \cos x \cdot (2\sin x \cdot \cos x)}{\sin^4 x} \]
06

Simplify the Expression

Simplify the expression obtained:\[ = \frac{-\sin x \cdot \sin^2 x - 2\cos^2 x \cdot \sin x}{\sin^4 x} = \frac{-\sin^3 x - 2\cos^2 x \cdot \sin x}{\sin^4 x} \] Factor out \(-\sin x\):\[ = \frac{-\sin x (\sin^2 x + 2\cos^2 x)}{\sin^4 x} \]
07

Final Result for Derivative

Further simplify the expression, knowing that the identity \( \sin^2 x + \cos^2 x = 1 \) yields:\[ = \frac{-\sin x (1 + \cos^2 x)}{\sin^4 x} \] \[ = \frac{-(1 + \cos^2 x)}{\sin^3 x} \]

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Key Concepts

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

Differentiation
Differentiation is a core concept in calculus focused on finding how a function changes at any given point. This change is represented by the derivative. In simple terms, differentiation measures how a function's output value changes as you slightly tweak its input value. For instance, if you have a function that describes the distance a car travels over time, the derivative of this function will give you the car's speed at any specific time.
To differentiate a function, you use certain rules and formulas. These rules help you tackle different function types, ensuring that you can always find the rate of change. In the context of this exercise, the main strategy involves breaking down the function and using the appropriate differentiation rules, like the Quotient Rule and Chain Rule, to find the solution.
Quotient Rule
The Quotient Rule is an essential tool in calculus for differentiating functions that are in the form of a fraction, where one function is divided by another. When faced with a quotient of two functions, say \( g(x) = \frac{u(x)}{v(x)} \), where both \( u(x) \) and \( v(x) \) are differentiable, the Quotient Rule provides a systematic way to find the derivative.
The formula is:
  • \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2} \)
It gives you the derivative of the entire expression by first differentiating each individual function. In our example:
- **Numerator** (\( u(x) = \cos x \)): The derivative is \( u'(x) = -\sin x \).
- **Denominator** (\( v(x) = \sin^2 x \)): Here, the Chain Rule is also used in differentiation since \( \sin^2 x \) is a composite function. The derivative is \( v'(x) = 2 \sin x \cdot \cos x \).
After calculating the derivatives of both functions, the Quotient Rule is applied to find the derivative of the original function, giving insight into how the function's output changes with input.
Trigonometric Functions
Trigonometric functions are functions that relate angles of a triangle to the lengths of its sides. These functions include sine (\( \sin \)), cosine (\( \cos \)), tangent (\( \tan \)), and their respective reciprocals. These functions are incredibly useful in mathematics and physics since they allow us to model patterns and phenomena like waves, circular motion, and oscillations.
In this exercise, we focus specifically on \( \cos x \) and \( \sin x \). Differentiating these functions requires understanding their properties:
  • The derivative of \( \cos x \) is \( -\sin x \).
  • The derivative of \( \sin x \) is \( \cos x \).
These properties are applied in calculus to find the rate at which these trigonometric functions change with respect to variables like angle or time.
This makes trigonometric derivatives a key aspect of calculus, helping solve a range of problems in sound waves, light waves, or simply any situation involving periodic patterns.

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