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Chapter 2: Problem 13

Find \(f^{\prime}(x)\) $$ f(x)=\frac{\cot x}{1+\csc x} $$

Short Answer

Expert verified
The derivative is \(f'(x) = \frac{-\csc^2 x - \csc^3 x + \csc x \cot^2 x}{(1 + \csc x)^2}\).

Step by step solution

01

Understand the Given Function

We are given the function \( f(x) = \frac{\cot x}{1 + \csc x} \). To differentiate it, we will need to apply the quotient rule because it is a fraction.
02

Recall the Quotient Rule

The quotient rule for differentiation is \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \). Here, \( u = \cot x \) and \( v = 1 + \csc x \). We will find their derivatives \( u' \) and \( v' \).
03

Differentiate \( u = \cot x \)

Find \( u' \):\( \cot x = \frac{\cos x}{\sin x} \), and its derivative is \( \frac{d}{dx}\cot x = -\csc^2 x \). Thus, \( u' = -\csc^2 x \).
04

Differentiate \( v = 1 + \csc x \)

Find \( v' \):\( \csc x = \frac{1}{\sin x} \), and its derivative is \( \frac{d}{dx}\csc x = -\csc x \cot x \). So, \( v' = -\csc x \cot x \).
05

Apply the Quotient Rule

Substitute \( u, v, u', v' \) into the quotient rule formula:\[ \frac{d}{dx} \left( \frac{\cot x}{1 + \csc x} \right) = \frac{-\csc^2 x (1 + \csc x) - (\cot x)(-\csc x \cot x)}{(1 + \csc x)^2} \]
06

Simplify the Expression

Start by expanding the terms in the numerator:\[ -\csc^2 x (1 + \csc x) + \cot x \cdot \csc x \cdot \cot x \]Simplify:\[ = -\csc^2 x - \csc^3 x + \csc x \cot^2 x \]The derivative is:\[ f'(x) = \frac{-\csc^2 x - \csc^3 x + \csc x \cot^2 x}{(1 + \csc x)^2} \]
07

Final Answer

The derivative of the function \( f(x) = \frac{\cot x}{1 + \csc x} \) is:\[ f'(x) = \frac{-\csc^2 x - \csc^3 x + \csc x \cot^2 x}{(1 + \csc x)^2} \]

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Key Concepts

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

Quotient Rule
When you need to differentiate a function that is a fraction, like \( \frac{\cot x}{1+\csc x} \), the quotient rule comes in handy. This method simplifies the process of finding derivatives of ratios of two functions.
The quotient rule formula is:
  • \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \)
Here, \( u \) is the numerator and \( v \) is the denominator. To find the derivative using this rule, follow these steps:
  • Differentiate the top function \( u \) to get \( u' \)
  • Differentiate the bottom function \( v \) to get \( v' \)
  • Substitute \( u \), \( v \), \( u' \), and \( v' \) into the formula above
  • Simplify the result
Exercise an extra degree of care in simplifying, as this step can get slightly tedious. The quotient rule is particularly useful when both the numerator and denominator are themselves functions, not just numbers.
Trigonometric Functions
Trigonometric functions like \( \sin x, \cos x, \tan x, \cot x, \csc x, \) and \( \sec x \) are fundamental in calculus. These functions often show up in differentiation and integration.
In our exercise, the main trigonometric functions we deal with are \( \cot x \) (cotangent) and \( \csc x \) (cosecant). Understanding these functions helps to navigate differentiation problems more smoothly:
  • \( \cot x = \frac{\cos x}{\sin x} \), and its derivative is \( -\csc^2 x \)
  • \( \csc x = \frac{1}{\sin x} \), and its derivative is \( -\csc x \cot x \)
Knowing these derivatives by heart enables you to apply the quotient rule effectively when these functions are involved. Trigonometric derivatives often need careful attention to detail due to the multiple functions involved in their expressions.
Derivative Calculation
Calculating derivatives is a core function in calculus, particularly to find the rate of change of a function. Let's follow this through with our problem, using the quotient rule.
Here's how to calculate the derivative of \( f(x) = \frac{\cot x}{1 + \csc x} \):
  • Identify \( u = \cot x \) and \( v = 1 + \csc x \)
  • Find the derivative of the numerator, \( u \), which satisfies \( u' = -\csc^2 x \)
  • Calculate the derivative of the denominator, \( v \), which leads to \( v' = -\csc x \cot x \)
  • Substitute these into the quotient rule formula: \[ \frac{d}{dx} \left( \frac{\cot x}{1 + \csc x} \right) = \frac{-\csc^2 x(1 + \csc x) + \cot x \cdot \csc x \cdot \cot x}{(1 + \csc x)^2} \]
  • Simplify further by expanding and simplifying the terms in the numerator
Ensuring each step is executed correctly can clarify complex problems, and though the calculation can be lengthy, it's systematic with practice. By doing this, you can better comprehend how different calculus rules apply to finding derivatives of tricky expressions.

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Most popular questions from this chapter

Find \(f^{\prime}(x)\) $$ f(x)=\sin \left(\frac{1}{x^{2}}\right) $$

Find \(f^{\prime}(x)\) $$ f(x)=\sqrt{4+\sqrt{3 x}} $$

Find the indicated derivative. $$ \frac{d}{d \omega}\left[a \cos ^{2} \pi \omega+b \sin ^{2} \pi \omega\right] \quad(a, b \text { constants }) $$

You are asked in these exercises to determine whether a piecewise-defined function \(f\) is differentiable at a value \(x=x_{0}\) where \(f\) is defined by different formulas on different sides of \(x_{0} .\) You may use without proof the following result, which is a consequence of the Mean-Value Theorem (discussed in Section \(4.8) .\) Theorem. Let \(f\) be continuous at \(x_{0}\) and suppose that \(\lim _{x \rightarrow x_{0}} f^{\prime}(x)\) exists. Then \(f\) is differentiable at \(x_{0},\) and \(f^{\prime}\left(x_{0}\right)=\lim _{x \rightarrow x_{0}} f^{\prime}(x) .\) $$ \begin{array}{l}{\text { Let } \quad f(x)=\left\\{\begin{array}{ll}{x^{2}-16 x,} & {x<9} \\ {\sqrt{x},} & {x \geq 9}\end{array}\right.} \\ {\text { Is } f \text { continuous at } x=9 \text { ? Determine whether } f \text { is differ- }} \\ {\text { entiable at } x=9 . \text { If so, find the value of the derivative }} \\ {\text { there. }}\end{array} $$

Find \(d y / d x\) $$ y=x^{3} \sin ^{2}(5 x) $$
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