From 67a97b6c529302a1801e5ba54454fda2680f766e Mon Sep 17 00:00:00 2001 From: Oscar Levin Date: Thu, 2 Oct 2025 11:49:03 -0600 Subject: [PATCH 1/4] add 2.3 exercises --- C-2.html | 279 +++++++++++++++++ sec-1-8-tan-line-approx.html | 550 +++++++++++++++++++++++++++++++++ source/chap-2.xml | 3 +- source/sec-2-3-prod-quot.xml | 580 +++++++++-------------------------- 4 files changed, 968 insertions(+), 444 deletions(-) create mode 100644 C-2.html create mode 100644 sec-1-8-tan-line-approx.html diff --git a/C-2.html b/C-2.html new file mode 100644 index 0000000..d08120a --- /dev/null +++ b/C-2.html @@ -0,0 +1,279 @@ + + + + + + + + + + + + + + + + +AC Computing Derivatives + + + + + + + + + + + + + + + + + + + + + + +Skip to main content
+

The Active Calculus Companion

+

Oscar Levin, Christina Safranski

+
+
+ +

+Chapter 2 Computing Derivatives +

+
🔗
+
+PrevTopNext +
+
+
+ +
+ + diff --git a/sec-1-8-tan-line-approx.html b/sec-1-8-tan-line-approx.html new file mode 100644 index 0000000..ea77d21 --- /dev/null +++ b/sec-1-8-tan-line-approx.html @@ -0,0 +1,550 @@ + + + + + + + + + + + + + + + + +AC The tangent line approximation + + + + + + + + + + + + + + + + + + + + + + +Skip to main content
+

The Active Calculus Companion

+

Oscar Levin, Christina Safranski

+
+
+ +

+Section 1.8 The tangent line approximation +

+

+Subsection 1.8.1 Additional Practice +

+

+Exercise 1.8.1. +

+
+
+
+
In the parts below, \(f{\small(x)}\) and \(f'{\small(x)}\) have values given in the table below
🔗
+
+ + + + + + + + + + + + + + + + + + + + + + + + + +
\(x\)\(f{\small(x)}\)\(f'{\small(x)}\)
\(1\)\(3\)\(4\)
\(2\)\(4\)\(3\)
\(3\)\(2\)\(1\)
\(4\)\(1\)\(2\)
Give formulas for the following tangent lines at \(x=1\text{.}\)
🔗
+
+(A) The tangent line to \(y=f{\small(x)}\) at \(x=1\text{.}\)
🔗
+
+\(\quad y =\)
🔗
+
+(B) The tangent line to \(y=f{\small(x)} + x\) at \(x=1\text{.}\)
🔗
+
+\(\quad y =\)
🔗
+
+(C) The tangent line to \(y= x f{\small(x)}\) at \(x=1\text{.}\)
🔗
+
+\(\quad y =\)
🔗
+
+(D) The tangent line to \(y= f{\small(x+1)}\) at \(x=1\text{.}\)
🔗
+
+\(\quad y =\)
🔗
+
+
+
+
🔗

+Exercise 1.8.2. +

+
+
+
+
Given that \(\small{f(8) = -5}\) and \(\small{f'(8) = 7}\text{,}\) find an equation for the tangent line to the graph of \(\small{y = f(x)}\) at \(\small{x = 8}\text{.}\)
🔗
+
+\(\small{y =}\)
🔗
+
+
+
+
🔗

+Exercise 1.8.3. +

+
+
+
+
Suppose \(f(x)\) is a function with \(f(3) = 2\) and \(f'(3) = -2\text{.}\)
🔗
+
+A. \(\quad\) Give the equation of the tangent line to \(y=f(x)\) at \(x=3\)
🔗
+
+\(\quad y =\)
🔗
+
+B. \(\quad\) Estimate the following values.
🔗
+
+\(\quad \displaystyle f\left({\frac{7}{2}}\right) \approx\)
🔗
+
+\(\quad \displaystyle f\left({\frac{3}{2}}\right) \approx\)
🔗
+
+
+
+
🔗

+Exercise 1.8.4. +

+
+
+
+
Given that \(f(-8) = 6\) and \(f'(-8) = 6\text{,}\) approximate \(f(-8.2)\text{.}\)
🔗
+
+\(f(-8.2) \approx\)
🔗
+
+
+
+
🔗

+Exercise 1.8.5. +

+
+
+
+
Given that \(P(88) = 29\) and \(P'(88) = 3\text{,}\) approximate \(P(94)\text{.}\)
🔗
+
+\(P(94) \approx\)
🔗
+
+
+
+
🔗

+Exercise 1.8.6. +

+
+
+
+
It turns out that the function \(f(x) = {\frac{1}{x}}\) has derivative \(f'(x) = {\frac{-1}{x^{2}}}\text{.}\)
🔗
+
Find the linear approximation \(L(x)\) to \(y={\frac{1}{x}}\) at \(x=-4\text{.}\)
🔗
+
+\(L(x)=\)
🔗
+
+
+
+
🔗

+Exercise 1.8.7. +

+
+
+
+
The function \(f(x) = x^{4}+3x^{2}+2\) has derivative \(f'(x) = 4x^{3}+6x\text{.}\)
🔗
+
Find the local linearization \(L(x)\) of \(f\) at \(x=2.\)
🔗
+
+\(L(x) =\)
🔗
+
+
+
+
🔗
🔗

+Subsection 1.8.2 Foundations +

+
To be able to find the equation of a tangent line, you must know how to find the equation of a line given a point and a slope. Here are some practice exercises for this foundational skill.
🔗
+

+Exercise 1.8.8. +

+
+
+
+
A line’s equation is given in point-slope form:
🔗
+
+\({y-4}={4\mathopen{}\left(x-2\right)}\)
🔗
+
This line’s slope is .
🔗
+
A point on this line that is apparent from the given equation is .
🔗
+
+
+
+
🔗

+Exercise 1.8.9. +

+
+
+
+
A line’s equation is given in point-slope form:
🔗
+
+\({y}={4\mathopen{}\left(x-4\right)+19}\)
🔗
+
This line’s slope is .
🔗
+
A point on this line that is apparent from the given equation is .
🔗
+
+
+
+
🔗

+Exercise 1.8.10. +

+
+
+
+
A line’s equation is given in point-slope form:
🔗
+
+\({y}={-5\mathopen{}\left(x+4\right)+17}\)
🔗
+
This line’s slope is .
🔗
+
A point on this line that is apparent from the given equation is .
🔗
+
+
+
+
🔗

+Exercise 1.8.11. +

+
+
+
+
Give the equations for the following lines:
🔗
+
+(A) Line through the point \((3,1)\) with slope \(-{\frac{5}{2}}\text{.}\)
🔗
+
+\(\qquad y =\)
🔗
+
+(B) Line satisfying \(y(-4) = 3\) and \(y(2) = 4\text{.}\)
🔗
+
+\(\qquad y =\)
🔗
+
+
+
+
🔗

+Exercise 1.8.12. +

+
+
+
+
Find an equation of the line passing through the points \((-8,3)\) with the slope \(m = {{\frac{5}{7}}}\)
🔗
+
+
  • +
    Do not use decimal approximations in your answer.
    🔗
    +
    +
    🔗
    +
+
🔗
+
+
Hint.
+
+Hint: Use the point-slope form of a line. \(y = m (x-x_A) + y_A\)
🔗
+
+
🔗
+
+
+
+
🔗

+Exercise 1.8.13. +

+
+
+
+
Find an equation for the line graphed below:
🔗
+
+
  • +
    Do not use decimal approximations in your answer.
    🔗
    +
    +
    🔗
    +
+
🔗
+
+
Hint.
+
Start by identifying a couple of “nice” points on the grid that the line passes through.
🔗
+
+
Once have two points for your line to pass through, you can determine the slope.
🔗
+
+
Use the point-slope form of a line. \(y = m (x-x_A) + y_A\)
🔗
+
+
🔗
+
+
+
+
🔗

+Exercise 1.8.14. +

+
+
+
+
Find the formula for \(f\text{.}\)
🔗
+
+\(f(x)=\)
🔗
+
+
+
+
🔗

+Exercise 1.8.15. +

+
+
+
+
Find the formula for \(f\text{.}\)
🔗
+
+\(f(x)=\)
🔗
+
+
+
+
🔗
🔗
🔗
+
+PrevTopNext +
+
+
+ +
+ + diff --git a/source/chap-2.xml b/source/chap-2.xml index e7fa499..cd2c150 100644 --- a/source/chap-2.xml +++ b/source/chap-2.xml @@ -5,7 +5,8 @@ Computing Derivatives - - Introduction -

- So far, we can differentiate power functions (x^n), - exponential functions (a^x), - and the two fundamental trigonometric functions (\sin(x) and \cos(x)). - With the sum rule and constant multiple rules, - we can also compute the derivative of combined functions. -

- - - -

- Find the derivative of - - f(x) = 7x^{11} - 4 \cdot 9^x + \pi \sin(x) - \sqrt{3}\cos(x) - . - Because f is a sum of basic functions, - we can now quickly say that f'(x) = 77x^{10} - 4 \cdot 9^x \ln(9) + \pi \cos(x) + \sqrt{3} \sin(x). -

- - -

- What about a product or quotient of two basic functions, such as - - p(z) = z^3 \cos(z) - , - or - - q(t) = \frac{\sin(t)}{2^t} - ? -

- -

- While the derivative of a sum is the sum of the derivatives, - it turns out that the rules for computing derivatives of products and quotients are more complicated. -

- - - - - - - - - The product rule -

- As part (b) of Preview Activity shows, - it is not true in general that the derivative of a product of two functions is the product of the derivatives of those functions. - To see why this is the case, - we consider an example involving meaningful functions. -

- -

- Say that an investor is regularly purchasing stock in a particular company. - Let N(t) represent the number of shares owned on day t, - where t = 0 represents the first day on which shares were purchased. - Let S(t) give the value of one share of the stock on day t; - note that the units on S(t) are dollars per share. - To compute the total value of the stock on day t, - we take the product - - V(t) = N(t) \, \text{shares} \cdot S(t) \, \text{dollars per share} - . - Observe that over time, - both the number of shares and the value of a given share will vary. - The derivative N'(t) measures the rate at which the number of shares is changing, - while S'(t) measures the rate at which the value per share is changing. - How do these respective rates of change affect the rate of change of the total value function? -

- -

- To help us understand the relationship among changes in N, - S, and V, let's consider some specific data. - -

    -
  • - Suppose that on day 100, - the investor owns 520 shares of stock and the stock's current value is $27.50 per share. - This tells us that N(100) = 520 and S(100) = 27.50. -
    • - -
  • - On day 100, - the investor purchases an additional 12 shares - (so the number of shares held is rising at a rate of 12 shares per day). -
    • - -
  • - On that same day the price of the stock is rising at a rate of 0.75 dollars per share per day. -
    • -
-

- -

- In calculus notation, the latter two facts tell us - that N'(100) = 12 - (shares per day) - and S'(100) = 0.75 - (dollars per share per day). - At what rate is the value of the investor's total holdings changing on day 100? -

- -

- Observe that the increase in total value comes from two sources: - the growing number of shares, - and the rising value of each share. - If only the number of shares is increasing - (and the value of each share is constant), - the rate at which total value would rise is the product of the current value of the shares and the rate at which the number of shares is changing. - That is, the rate at which total value would change is given by - - S(100) \cdot N'(100) = 27.50 \, \frac{\text{dollars} }{\text{share} } \cdot 12 \, \frac{\text{shares} }{\text{day} } = 330 \, \frac{\text{dollars} }{\text{day} } - . -

- -

- Note particularly how the units make sense and show the rate at which the total value V is changing, - measured in dollars per day. -

- -

- If instead the number of shares is constant, - but the value of each share is rising, - the rate at which the total value would rise is the product of the number of shares and the rate of change of share value. - The total value is rising at a rate of - - N(100) \cdot S'(100) = 520 \, \text{shares} \cdot 0.75 \, \frac{\text{dollars per share} }{\text{day} } = 390 \, \frac{\text{dollars} }{\text{day} } - . -

- -

- Of course, when both the number of shares and the value of each share are changing, - we have to include both of these sources. - In that case the rate at which the total value is rising is - - V'(100) = S(100) \cdot N'(100) + N(100) \cdot S'(100) = 330 + 390 = 720 \, \frac{\text{dollars} }{\text{day} } - . -

- -

- We expect the total value of the investor's holdings to rise by about $720 on the 100th day. - While this example highlights why the product rule is true, - there are some subtle issues to recognize. - For one, if the stock's value really does rise exactly $0.75 on day 100, - and the number of shares really rises by 12 on day 100, - then we'd expect that V(101) = N(101) \cdot S(101) = 532 \cdot 28.25 = 15029. - If, as noted above, - we expect the total value to rise by $720, - then with V(100) = N(100) \cdot S(100) = 520 \cdot 27.50 = 14300, - then it seems we should find that V(101) = V(100) + 720 = 15020. - Why do the two results differ by 9? - One way to understand why this difference occurs is to recognize that N'(100) = 12 represents an - instantaneous rate of change, - while our (informal) discussion has also thought of this number as the total change in the number of shares over the course of a single day. - The formal proof of the product rule reconciles this issue by taking the limit as the change in the input tends to zero. -

- -

- Next, we expand our perspective from the specific example above to the more general and abstract setting of a product P of two differentiable functions, - f and g. - If P(x) = f(x) \cdot g(x), - our work above suggests that P'(x) = f(x) g'(x) + g(x) f'(x). - Indeed, a formal proof using the limit definition of the derivative can be given to show that the following rule, - called the product rule, - product rule - holds in general. -

- - - The Product Rule -

- product rule - If f and g are differentiable functions, - then their product P(x) = f(x) \cdot g(x) is also a differentiable function, and - - P'(x) = f(x) g'(x) + g(x) f'(x) - . -

- - -

- In light of the earlier example involving shares of stock, - the product rule also makes sense intuitively: - the rate of change of P should take into account both how fast f and g are changing, - as well as how large f and g are at the point of interest. - In words the product rule says: - if P is the product of two functions f - (the first function) - and g - (the second), - then the derivative of P is the first times the derivative of the second, - plus the second times the derivative of the first. - It is often a helpful mental exercise to say this phrasing aloud when executing the product rule. -

- - - -

- Use the product rule to differentiate P(z) = z^3 \cdot \cos(z). -

- - -

- In P(z) = z^3 \cdot \cos(z), - the first function is z^3 and the second function is \cos(z). - By the product rule, P' will be given by the first, - z^3, - times the derivative of the second, - -\sin(z), plus the second, - \cos(z), times the derivative of the first, 3z^2. - That is, - - P'(z) = z^3(-\sin(z)) + \cos(z) 3z^2 = -z^3 \sin(z) + 3z^2 \cos(z) - . -

- - - - - - - - The quotient rule -

- Because quotients and products are closely linked, - we can use the product rule to understand how to take the derivative of a quotient. - Let Q(x) be defined by Q(x) = f(x)/g(x), - where f and g are both differentiable functions. - It turns out that Q is differentiable everywhere that g(x) \ne 0. - We would like a formula for Q' in terms of f, - g, f', and g'. - - Multiplying both sides of the formula Q = f/g by g, - we observe that - - f(x) = Q(x) \cdot g(x) - . -

- -

- Applying the product rule to differentiate f, it follows - - f'(x) = Q(x) g'(x) + g(x) Q'(x) - . -

- -

- Since we want to know a formula for Q', - we solve this most recent equation for Q'(x) and find - - Q'(x) g(x) = f'(x) - Q(x) g'(x) - . - Dividing both sides by g(x), we have - - Q'(x) = \frac{f'(x) - Q(x) g'(x)}{g(x)} - . -

- -

- Finally, we recall that Q(x) = \frac{f(x)}{g(x)}. - Substituting this expression in the preceding equation and simplifying, it follows that - - Q'(x) \amp= \frac{f'(x) - \frac{f(x)}{g(x)} g'(x)}{g(x)} - \amp= \frac{f'(x) - \frac{f(x)}{g(x)} g'(x)}{g(x)} \cdot \frac{g(x)}{g(x)} - \amp= \frac{g(x) f'(x) - f(x) g'(x)}{g(x)^2} - . -

- -

- The preceding argument results in the - quotient rule. -

- - - The Quotient Rule -

- quotient rule - If f and g are differentiable functions, - then their quotient Q(x) = \frac{f(x)}{g(x)} is also a differentiable function for all x where g(x) \ne 0 and - - Q'(x) = \frac{g(x) f'(x) - f(x) g'(x)}{g(x)^2} - . -

- - -

- As with the product rule, - it can be helpful to think of the quotient rule verbally. - If a function Q is the quotient of a top function f and a bottom function g, - then Q' is given by the bottom times the derivative of the top, - minus the top times the derivative of the bottom, - all over the bottom squared. -

- - - -

- Use the quotient rule to differentiate Q(t) = \dfrac{\sin(t)}{2^t}. -

- - -

- Since Q(t) = \frac{\sin(t)}{2^t}, - we call \sin(t) the top function and 2^t the bottom function. - By the quotient rule, - Q' is given by the bottom, - 2^t, - times the derivative of the top, \cos(t), - minus the top, \sin(t), - times the derivative of the bottom, - 2^t \ln(2), all over the bottom squared, (2^t)^2. - That is, - - Q'(t) = \frac{2^t \cos(t) - \sin(t) 2^t \ln(2)}{(2^t)^2} - . -

- -

- In this particular example, - it is possible to simplify Q'(t) by removing a factor of 2^t from both the numerator and denominator, - so that - - Q'(t) = \frac{\cos(t) - \sin(t) \ln(2)}{2^t} - . -

- - - -

- In general, - we must be careful in doing any such simplification, - as we don't want to execute the quotient rule correctly but then make an algebra error. -

- - - - - - Combining rules -

- In order to apply the derivative shortcut rules correctly we must recognize the fundamental structure of a function. -

- - - -

- Determine the derivative of the function - - f(x) = x\sin(x) + \frac{x^2}{\cos(x) + 2} - . - Clearly state which derivative rules you use and how they were applied. -

- - -

- To differentiate any complicated function, our first task is to recognize the structure of the function. - This function f is a sum of two slightly less complicated functions, - so we can apply the sum rule - When taking a derivative that involves the use of multiple derivative rules, - it is often helpful to use the notation - \frac{d}{dx} \left[ ~~\right] to wait to apply subsequent rules. - This is demonstrated both in this example and the one that follows. - to get - - f'(x) \amp= \frac{d}{dx} \left[ x\sin(x) + \frac{x^2}{\cos(x) + 2} \right] - \amp= \frac{d}{dx} \left[ x\sin(x) \right] + \frac{d}{dx}\left[ \frac{x^2}{\cos(x) + 2} \right] - -

- -

- Now, the left-hand term above is a product, - so the product rule is needed there, - while the right-hand term is a quotient, - so the quotient rule is required. - Applying these rules respectively, we find that - - f'(x) \amp= \left( x \cos(x) + \sin(x) \right) + \frac{(\cos(x) + 2) 2x - x^2(-\sin(x))}{(\cos(x) + 2)^2} - \amp= x \cos(x) + \sin(x) + \frac{2x\cos(x) + 4x + x^2\sin(x)}{(\cos(x) + 2)^2} - . -

- - - - - -

- Determine the derivative of the function - - s(y) = \frac{y \cdot 7^y}{y^2 + 1} - . - Clearly state which rules you used and how they were applied. -

- - -

- The function s is a quotient of two simpler functions, - so the quotient rule will be needed. - To begin, we set up the quotient rule and use the notation - \frac{d}{dy} to indicate the derivatives of the numerator and denominator. - Thus, - - s'(y) = \frac{(y^2 + 1) \cdot \frac{d}{dy}\left[ y \cdot 7^y \right] - y \cdot 7^y \cdot \frac{d}{dy}\left[y^2 + 1 \right]}{(y^2 + 1)^2} - . - Now, there remain two derivatives to determine. - The first one, - \frac{d}{dy}\left[ y \cdot 7^y \right] calls for use of the product rule, - while the second, - \frac{d}{dy}\left[y^2 + 1 \right] needs only the sum rule. - Applying these rules, we now have - - s'(y) = \frac{(y^2 + 1) [y \cdot 7^y \ln(7) + 7^y \cdot 1] - y \cdot 7^y [2y]}{(y^2 + 1)^2} - . - While some minor simplification is possible, - we are content to leave s'(y) in its current form. -

- - - -

- Success in applying derivative rules begins with recognizing the structure of the function, - followed by the careful and diligent application of the relevant derivative rules. - The best way to become proficient at this process is to do a large number of examples. -

- - - -

- As the algebraic complexity of the functions we are able to differentiate continues to increase, - it is important to remember that all of the derivative's meaning continues to hold. - Regardless of the structure of the function f, - the value of f'(a) tells us the instantaneous rate of change of f with respect to x at the moment x = a, - as well as the slope of the tangent line to - y = f(x) at the point (a,f(a)). -

- - - - Summary + Introduction

+ Active Calculus provides the following summary of section 2.3:

  • @@ -520,7 +84,137 @@ - - - + + Additional Practice + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + From 9d0e8ebf4468fa0e76b45e0152276c25d94570d0 Mon Sep 17 00:00:00 2001 From: Oscar Levin Date: Thu, 2 Oct 2025 11:57:32 -0600 Subject: [PATCH 2/4] reorganize sections for new structure --- source/sec-1-8-tan-line-approx.xml | 77 +++++++++++------------ source/sec-2-1-elem-rules.xml | 99 +++++++++++++++--------------- source/sec-2-2-sin-cos.xml | 18 +++--- source/sec-2-3-prod-quot.xml | 55 ++--------------- 4 files changed, 106 insertions(+), 143 deletions(-) diff --git a/source/sec-1-8-tan-line-approx.xml b/source/sec-1-8-tan-line-approx.xml index ad8571c..11860d3 100644 --- a/source/sec-1-8-tan-line-approx.xml +++ b/source/sec-1-8-tan-line-approx.xml @@ -73,44 +73,6 @@

    --> - - Additional Practice - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Foundations @@ -155,5 +117,44 @@ + + Additional Practice + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + diff --git a/source/sec-2-1-elem-rules.xml b/source/sec-2-1-elem-rules.xml index 967479c..d7d7b89 100644 --- a/source/sec-2-1-elem-rules.xml +++ b/source/sec-2-1-elem-rules.xml @@ -88,8 +88,56 @@

    --> - - Additional Practice + + Foundations +

    + Sometimes before computing a derivative, it is helpful to rewrite the function in a more standard form. Let's review how to write radicals as power functions, as well as powers of x in the denominator of a fraction as power functions. +

    + +

    + We start with some practice with basic rules of exponents. +

    + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +     + + + + Additional Practice Exercises @@ -145,52 +193,7 @@ -     - - - Foundations -

    - Sometimes before computing a derivative, it is helpful to rewrite the function in a more standard form. Let's review how to write radicals as power functions, as well as powers of x in the denominator of a fraction as power functions. -

    - -

    - We start with some practice with basic rules of exponents. -

    - - - - - - - - + - - - - - - - - - - - - - - - - - - - - - - - - - - -     diff --git a/source/sec-2-2-sin-cos.xml b/source/sec-2-2-sin-cos.xml index e27d1ed..419fa65 100644 --- a/source/sec-2-2-sin-cos.xml +++ b/source/sec-2-2-sin-cos.xml @@ -62,8 +62,14 @@

    --> - - Additional Practice + + Foundations + + + + + + Additional Practice Exercises @@ -88,11 +94,7 @@ - + + - - Foundations - - - diff --git a/source/sec-2-3-prod-quot.xml b/source/sec-2-3-prod-quot.xml index 4d37bc2..1b56938 100644 --- a/source/sec-2-3-prod-quot.xml +++ b/source/sec-2-3-prod-quot.xml @@ -34,58 +34,15 @@ - Introduction + Foundational Exercises

    - Active Calculus provides the following summary of section 2.3: -

      -
    • -

      - If a function is a sum, product, - or quotient of simpler functions, - then we can use the sum, product, - or quotient rules to differentiate it in terms of the simpler functions and their derivatives. -

      -
      • - -
    • -

      - The product rule tells us that if P is a product of differentiable functions f and g according to the rule P(x) = f(x) g(x), then - - P'(x) = f(x)g'(x) + g(x)f'(x) - . -

      -
      • - -
    • -

      - The quotient rule tells us that if Q is a quotient of differentiable functions f and g according to the rule Q(x) = \frac{f(x)}{g(x)}, then - - Q'(x) = \frac{g(x)f'(x) - f(x)g'(x)}{g(x)^2} - . -

      -
      • - -
    • -

      - Along with the constant multiple and sum rules, - the product and quotient rules enable us to compute the derivative of any function that consists of sums, - constant multiples, products, - and quotients of basic functions. - For instance, if F has the form - - F(x) = \frac{2a(x) - 5b(x)}{c(x) \cdot d(x)} - , - then F is a quotient, - in which the numerator is a sum of constant multiples and the denominator is a product. - This, the derivative of F can be found by applying the quotient rule and then using the sum and constant multiple rules to differentiate the numerator and the product rule to differentiate the denominator. -

      -
      • -

    + Coming soon. +

    - - Additional Practice + + Additional Practice Exercises @@ -214,7 +171,7 @@ - + From 9f44fac458b7a0fafaa1f681989726d019c4f3ab Mon Sep 17 00:00:00 2001 From: Oscar Levin Date: Thu, 2 Oct 2025 11:58:13 -0600 Subject: [PATCH 3/4] increase toc level --- publication/publication-rs.ptx | 2 +- publication/publication.ptx | 2 +- 2 files changed, 2 insertions(+), 2 deletions(-) diff --git a/publication/publication-rs.ptx b/publication/publication-rs.ptx index 4db5d5d..344d071 100644 --- a/publication/publication-rs.ptx +++ b/publication/publication-rs.ptx @@ -15,7 +15,7 @@ - + diff --git a/publication/publication.ptx b/publication/publication.ptx index b9a5228..1d66e22 100644 --- a/publication/publication.ptx +++ b/publication/publication.ptx @@ -15,7 +15,7 @@ - + From 86ebfdc0b9deb64908325427968d31b9c2059502 Mon Sep 17 00:00:00 2001 From: Oscar Levin Date: Thu, 2 Oct 2025 12:00:28 -0600 Subject: [PATCH 4/4] add empty 2.4 --- source/sec-2-4-other-trig.xml | 213 ++-------------------------------- 1 file changed, 9 insertions(+), 204 deletions(-) diff --git a/source/sec-2-4-other-trig.xml b/source/sec-2-4-other-trig.xml index bfcd898..814cbec 100644 --- a/source/sec-2-4-other-trig.xml +++ b/source/sec-2-4-other-trig.xml @@ -3,7 +3,7 @@
    Derivatives of other trigonometric functions - + - Introduction + + Foundations

    - One of the powerful themes in trigonometry - trigonometry - comes from a very simple idea: - locating a point on the unit circle. -

    - -
    - The unit circle and the definition of the sine and cosine functions. - -
    - -

    - Because each angle \theta in standard position corresponds to one and only one point (x,y) on the unit circle, - the x- and y-coordinates of this point are each functions of \theta. - In fact, this is the very definition of - \cos(\theta) and \sin(\theta): - \cos(\theta) is the x-coordinate of the point on the unit circle corresponding to the angle \theta, - and \sin(\theta) is the y-coordinate. - From this simple definition, - all of trigonometry is founded. - For instance, the Fundamental Trigonometric Identity, - trigonometryfundamental trigonometric identity - - \sin^2(\theta) + \cos^2(\theta) = 1 - , - is a restatement of the Pythagorean Theorem, - applied to the right triangle shown in Figure. -

    - -

    - There are four other trigonometric functions, - each defined in terms of the sine and/or cosine functions. - -

      -
    • - The tangent function - tangent - is defined by \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}; -
      • - -
    • - the cotangent function is its reciprocal: - \cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}. -
      • - -
    • - The secant function is the reciprocal of the cosine function, - \sec(\theta) = \frac{1}{\cos(\theta)}; -
      • - -
    • - and the cosecant function is the reciprocal of the sine function, - \csc(\theta) = \frac{1}{\sin(\theta)}. -
      • -
    - - These six trigonometric functions together offer us a wide range of flexibility in problems involving right triangles. -

    - -

    - Because we know the derivatives of the sine and cosine function, - we can now develop shortcut differentiation rules for the tangent, - cotangent, - secant, and cosecant functions. - In this section's preview activity, - we work through the steps to find the derivative of y = \tan(x). -

    - - - -     - - - Derivatives of the cotangent, secant, and cosecant functions - cotangent - secant - cosecant - -

    - In Preview Activity, - we found that the derivative of the tangent function can be expressed in several ways, - but most simply in terms of the secant function. - Next, we develop the derivative of the cotangent function. -

    - -

    - Let g(x) = \cot(x). - To find g'(x), we observe that - g(x) = \frac{\cos(x)}{\sin(x)} and apply the quotient rule. - Hence - - g'(x) =\mathstrut \amp \frac{\sin(x)(-\sin(x)) - \cos(x) \cos(x)}{\sin^2(x)} - =\mathstrut \amp -\frac{\sin^2(x) + \cos^2(x)}{\sin^2(x)} - -

    - -

    - By the Fundamental Trigonometric Identity, - we see that g'(x) = -\frac{1}{\sin^2(x)}, and - recalling that \csc(x) = \frac{1}{\sin(x)}, - it follows that we can express g' by the rule - - g'(x) = -\csc^2(x) - . -

    - -

    - Note that neither g nor g' is defined when \sin(x) = 0, - which occurs at every integer multiple of \pi. - Hence we have the following rule. -

    - - - The derivative of the cotangent function -

    - derivativecotangent - For all real numbers x such that x \ne k\pi, - where k = 0, \pm 1, \pm 2, \ldots, - - \frac{d}{dx} [\cot(x)] = -\csc^2(x) - . -

    -     - -

    - Notice that the derivative of the cotangent function is very similar to the derivative of the tangent function we discovered in Preview Activity. -

    - - - The derivative of the tangent function -

    - derivativetangent - For all real numbers x such that x \ne \frac{(2k+1)\pi}{2}, - where k = \pm 1, \pm 2, \ldots, - - \frac{d}{dx} [\tan(x)] = \sec^2(x) - . -

    -     - -

    - In the next two activities, - we develop the rules for differentiating the secant and cosecant functions. -

    - - - - - -

    - Using the quotient rule we have determined the derivatives of the tangent, - cotangent, - secant, and cosecant functions, - expanding our overall library of functions we can differentiate. - Observe that just as the derivative of any polynomial function is a polynomial, - and the derivative of any exponential function is another exponential function, - so it is that the derivative of any basic trigonometric function is another function that consists of basic trigonometric functions. - This makes sense because all trigonometric functions are periodic, - and hence their derivatives will be periodic, too. -

    - -

    - The derivative retains all of its fundamental meaning as an instantaneous rate of change and as the slope of the tangent line to the function under consideration. -

    - - -     - - - Summary -

    -

      -
    • -

      - The derivatives of the other four trigonometric functions are - - \frac{d}{dx}[\tan(x)] = \sec^2(x), \ \ \frac{d}{dx}[\cot(x)] = -\csc^2(x) - , - - \frac{d}{dx}[\sec(x)] = \sec(x)\tan(x), \ \text{and} \ \frac{d}{dx}[\csc(x)] = -\csc(x)\cot(x) - . - Each derivative exists and is defined on the same domain as the original function. - For example, - both the tangent function and its derivative are defined for all real numbers x such that x \ne \frac{k\pi}{2}, - where k = \pm 1, \pm 2, \ldots. -

      -
      • - -
    • -

      - The four rules for the derivatives of the tangent, - cotangent, secant, - and cosecant can be used along with the rules for power functions, - exponential functions, and the sine and cosine, - as well as the sum, constant multiple, - product, and quotient rules, - to quickly differentiate a wide range of different functions. -

      -
      • -
    + Coming soon.

    - + + Additional Practice + +