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Chapter 3: Problem 35

Find the derivatives of the functions in Exercises \(19-40\) $$f(\theta)=\left(\frac{\sin \theta}{1+\cos \theta}\right)^{2}$$

Short Answer

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

Step by step solution

01

Identify the outer function

The function to be differentiated is expressed as a power function: \( f(\theta) = \left( \frac{\sin \theta}{1+\cos \theta} \right)^2 \). Here, the outer function is represented by \( u^2 \) where \( u = \frac{\sin \theta}{1+\cos \theta} \). We will need the chain rule to differentiate this.
02

Differentiate the outer function

Differentiate the outer function \( y = u^2 \) with respect to \( u \). The derivative is \( \frac{d}{du}(u^2) = 2u \).
03

Identify the inner function

The inner function is \( u = \frac{\sin \theta}{1+\cos \theta} \). We will differentiate this expression to find \( \frac{du}{d\theta} \).
04

Differentiate the inner function

Use the quotient rule which states that if \( g(\theta) = \frac{a(\theta)}{b(\theta)} \), then \( g'(\theta) = \frac{a'(\theta) b(\theta) - a(\theta) b'(\theta)}{(b(\theta))^2} \). Let \( a(\theta) = \sin \theta \) and \( b(\theta) = 1 + \cos \theta \) with \( a'(\theta) = \cos \theta \) and \( b'(\theta) = -\sin \theta \).
05

Apply the quotient rule to the inner function

Using the quotient rule, the derivative of the inner function is:\[\frac{du}{d\theta} = \frac{\cos \theta (1 + \cos \theta) - \sin \theta (-\sin \theta)}{(1 + \cos \theta)^2}\]Simplify this to:\[\frac{du}{d\theta} = \frac{\cos \theta + \cos^2 \theta + \sin^2 \theta}{(1 + \cos \theta)^2}\]Since \( \sin^2 \theta + \cos^2 \theta = 1 \), it further simplifies to:\[\frac{du}{d\theta} = \frac{\cos \theta + 1}{(1 + \cos \theta)^2}\]
06

Apply the chain rule

According to the chain rule, the derivative of \( f(\theta) \) is the derivative of the outer function multiplied by the derivative of the inner function:\[\frac{df}{d\theta} = 2u \cdot \frac{du}{d\theta}\]Substitute \( u = \frac{\sin \theta}{1+\cos \theta} \) and simplify:\[\frac{df}{d\theta} = 2 \left( \frac{\sin \theta}{1+\cos \theta} \right) \cdot \frac{\cos \theta + 1}{(1 + \cos \theta)^2}\]This simplifies to:\[\frac{df}{d\theta} = \frac{2 \sin \theta (\cos \theta + 1)}{(1 + \cos \theta)^3}\]

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Key Concepts

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

Chain Rule
The chain rule is a fundamental tool in calculus when dealing with composite functions. These are functions inside another function. To apply the chain rule, you first identify the "outer" and "inner" functions. In the given example, we have the function \( f(\theta) = \left( \frac{\sin \theta}{1+\cos \theta} \right)^2 \). Here, the outer function is \( u^2 \) and the inner function is \( u = \frac{\sin \theta}{1+\cos \theta} \).
To differentiate using the chain rule, follow these steps:
  • Differentiate the outer function with respect to the inner function: for \( y = u^2 \), the derivative \( \frac{d}{du}(u^2) = 2u \).
  • Next, differentiate the inner function \( u \) with respect to the original variable \( \theta \), which involves using the quotient rule.
  • Multiply the derivative of the outer function by the derivative of the inner function to get the derivative of the composite function.
This rule simplifies the process of differentiating complex functions, helping break them into more manageable parts.
Quotient Rule
The quotient rule is important when the function you need to differentiate is a ratio of two differentiable functions. The quotient rule states:\[g'(\theta) = \frac{a'(\theta) b(\theta) - a(\theta) b'(\theta)}{(b(\theta))^2}\]where \( a(\theta) \) is the numerator and \( b(\theta) \) is the denominator.

In the exercise, the inner function \( u = \frac{\sin \theta}{1+\cos \theta} \) uses the quotient rule for differentiation. Here:
  • \( a(\theta) = \sin \theta \) and \( a'(\theta) = \cos \theta \)
  • \( b(\theta) = 1 + \cos \theta \) and \( b'(\theta) = -\sin \theta \)
  • Apply the quotient rule and simplify: \( \frac{du}{d\theta} = \frac{\cos \theta (1 + \cos \theta) - \sin \theta (-\sin \theta)}{(1 + \cos \theta)^2} \)
  • Using \( \sin^2 \theta + \cos^2 \theta = 1 \), the expression simplifies further to \( \frac{\cos \theta + 1}{(1 + \cos \theta)^2} \).
The quotient rule is essential to handle differentiating fractions of functions efficiently.
Trigonometric Functions
Trigonometric functions like \( \sin \theta \) and \( \cos \theta \) are frequently encountered in calculus. A strong grasp of their properties is necessary for differentiation and integration. The derivatives of these basic trigonometric functions are essential:
  • \( (\sin \theta)' = \cos \theta \)
  • \( (\cos \theta)' = -\sin \theta \)
These properties were directly used in the quotient rule to find the derivative of \( u = \frac{\sin \theta}{1+\cos \theta} \).
Understanding and remembering these derivatives simplifies calculations in calculus, especially when combined with other rules like the chain and quotient rules. Knowing when and how to apply these derivatives can aid significantly in solving complex derivative problems smoothly.

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Most popular questions from this chapter

a. Find equations for the tangents to the curves \(y=\sin 2 x\) and \(y=-\sin (x / 2)\) at the origin. Is there anything special about how the tangents are related? Give reasons for your answer. b. Can anything be said about the tangents to the curves \(y=\sin m x\) and \(y=-\sin (x / m)\) at the origin \((m\) a constant \(\neq 0) ?\) Give reasons for your answer. c. For a given \(m,\) what are the largest values the slopes of the curves \(y=\sin m x\) and \(y=-\sin (x / m)\) can ever have? Give reasons for your answer. d. The function \(y=\sin x\) completes one period on the interval \([0,2 \pi],\) the function \(y=\sin 2 x\) completes two periods, the function \(y=\sin (x / 2)\) completes half a period, and so on. Is there any relation between the number of periods \(y=\sin m x\) completes on \([0,2 \pi]\) and the slope of the curve \(y=\sin m x\)

Use a CAS to perform the following steps. \begin{equation} \begin{array}{l}{\text { a. Plot the equation with the implicit plotter of a CAS. Check to }} \\ {\text { see that the given point } P \text { satisfies the equation. }} \\ {\text { b. Using implicit differentiation, find a formula for the deriva- }} \\ {\text { tive } d y / d x \text { and evaluate it at the given point } P .}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { c. Use the slope found in part (b) to find an equation for the tan- }} \\ {\text { gent line to the curve at } P \text { . Then plot the implicit curve and }} \\ {\text { tangent line together on a single graph. }}\end{array} \end{equation} \begin{equation} x y^{3}+\tan (x+y)=1, \quad P\left(\frac{\pi}{4}, 0\right) \end{equation}

Use a CAS to perform the following steps for the functions \begin{equation} \begin{array}{l}{\text { a. Plot } y=f(x) \text { to see that function's global behavior. }} \\ {\text { b. Define the difference quotient } q \text { at a general point } x, \text { with }} \\ {\text { general step size } h .} \\\ {\text { c. Take the limit as } h \rightarrow 0 . \text { What formula does this give? }} \\ {\text { d. Substitute the value } x=x_{0} \text { and plot the function } y=f(x)} \\ \quad {\text { together with its tangent line at that point. }} \\ {\text { e. Substitute various values for } x \text { larger and smaller than } x_{0} \text { into }} \\ \quad {\text { the formula obtained in part (c). Do the numbers make sense }} \\ \quad {\text { with your picture? }} \\ {\text { f. Graph the formula obtained in part (c). What does it mean }} \\ \quad {\text { when its values are negative? Zero? Positive? Does this make }} \\ \quad {\text { sense with your plot from part (a)? Give reasons for your }} \\ \quad {\text { answer. }}\end{array} \end{equation} $$f(x)=x^{1 / 3}+x^{2 / 3}, \quad x_{0}=1$$

Moving along a parabola A particle moves along the parabola \(y=x^{2}\) in the first quadrant in such a way that its \(x\)-coordinate (measured in meters) increases at a steady 10 \(\mathrm{m} / \mathrm{sec} .\) How fast is the angle of inclination \(\theta\) of the line joining the particle to the origin changing when \(x=3 \mathrm{m} ?\)

a. Find the tangent to the curve \(y=2 \tan (\pi x / 4)\) at \(x=1\) b. Slopes on a tangent curve What is the smallest value the slope of the curve can ever have on the interval \(\quad-2
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