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# 4.2: Derivatives

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## Tangent Lines and Rates of Change

CONTENT

With a careful enough picture, the concept here can be made crystal clear. The goal in making such a picture should be to draw a large enough diagram with as little clutter as possible. Here is an example:

This picture, carefully explained, will make it easier for any student to see that the slope of the secant line is just $\frac{f(x+h)-f(x)}{h}$ and that as $h \rightarrow 0$ the secant approaches the tangent at $x$.

PROCESS

The best mechanism to give everyone a chance to understand this is by working carefully with the geometry. That is, the students should all work one simple example in lots of detail. Give them all three big sheets of graph paper and have them graph the function $y=x^2$ between $x=0$ and $x=5$, but using the entire sheet so that they can see it zoomed in very closely. They should do this carefully on each sheet so that each contains a careful copy of the plot below:

Then, students should draw in the secant line through $(1,1)$ and $(1+3,(1+3)^2 )=(4,16)$ on one of their pages:

Students should find the slope of this line and show their work as follows:

$m = \frac{16-1}{4-1} = \frac{15}{3} = 5$

Next, students should repeat this process for the secant line through $(1,1)$ and $(1+2,(1+2)^2 )=(3,9)$. The diagram will look like:

and the slope calculation should look like:

$m = \frac{9-1}{3-1} = \frac{8}{2} = 4$

and again now for the secant line through $(1,1)$ and $(1+.5,(1+.5)^2 )=(1.5,2.25)$. The diagram will look like:

and the calculation of slope will be:

$m = \frac{2.25 - 1}{1.5 - 1} = \frac{1.25}{0.5} = 2.5$

Finally, students should be asked to consider the general secant line through the points $(1,1)$ and $(1+h,(1+h)^2 )$, and find its slope:

$m = \frac{(1 + h)^2 - 1}{(1 + h) - 1} = \frac{1 + 2h + h^2 - 1}{h} = 2 + h$

Then ask students to consider how this equation works in terms of the calculations above. In the first diagram, we took points that were $h=3$ apart, and the slope of the secant was $5=2+h$. Then in the second diagram we took points that were $h=2$ apart and the slope of the secant was $4=2+h$. Then in the final diagram $h=0.5$ and slope was $2.5=2+h$. Clearly, as $h \rightarrow 0$ the slope goes to $2$. This is the slope of the tangent line at $x=2!$

PRODUCTS

The best way to test this material is to have students perform the above approximations for variety of functions and points in those functions. As intuition grows, students can be asked what they think the slope of the tangent should be ahead of time at a minimum or a maximum or, for example, how a plot of the tangent slope (derivative) will approximately look. Students should be given problems which apply these concepts to applications. For example:

i) A cannonball flies through the air in a parabolic shape. The height $y(t)$ satisfies the equation $y=5-4.9t^2$ and the horizontal distance $x(t)$ satisfies the equation $x=5t$.

a. Plot the horizontal and vertical positions as functions of time.

b. Draw in the secant line for each between $t=1$ and $t=3$.

c. Find the average vertical and horizontal velocities between $t=1$ and $t=3$

d. Using the same process, find the average vertical and horizontal velocities between $t=1$ and $t=1.5$

e. Draw the tangent lines at $t=1$ to each plot and find their slope $=$ the instantaneous velocities at $t=1$

f. Find the instantaneous velocities as functions of time throughout the entire flight

## The Derivative

CONTENT

Just like the previous section, this material is most accessibly introduced with a careful diagram like:

By making the definition of $f'(x)$ geometrically intuitive, it is much more easily recalled and manipulated.

It should be pointed out carefully that by understanding this picture and the very intuitive nature of tangent lines, differentiability is easy to understand. Simply put, a derivative exists wherever we can easily tell what the tangent line should look like. For example, at a corner like in the diagram below, the slope appears to jump instantly from one value to another. For this reason we cannot see exactly how the tangent should be placed at the corner.

The other way in which differentiability can fail is if the slope is infinite (that is, the tangent line is vertical) or the function is not continuous at the point of interest.

PROCESS

To teach this concept it is probably best to have students look at a number of functions, plot them, and assess where they are differentiable and then finding the derivative at these points by taking limits. Making this material interesting for students is a challenge, but it may be helpful to point out that this is the essence of calculus. If students can fully grasp the equation for a derivative and its geometric basis then the rest will come very easily. In other words, if learning calculus is like climbing a mountain then understanding the derivative is like the sheerest part of the climb. It is difficult but worthwhile to really work on getting the details straight here.

Differentiating the process here could be accomplished by having students divide into more or less equal groups or three. One student in the group will be asked to “teach” the other two exactly what continuity means. One student should be asked to “teach” exactly what a derivative is geometrically and analytically. And the final student should be asked to explain carefully the difference between continuity and differentiability.

PRODUCTS

Students should be tested on the concept of derivatives by asking questions that lead them towards the coming material. For example, the following are great questions:

1. Use the definition of the derivative to find $f'(x)$ if $f(x)=14$
2. Now generalize this, so that if $f(x)=C$ for any constant $C$ then $f'(x)$ is?
3. Use the definition of the derivative to find $f'(x)$ if $f(x)=x$
4. Use the definition of the derivative to find $f'(x)$ if $f(x)=x^2$
5. Use the definition of the derivative to find $f'(x)$ if $f(x)=x^3$
6. Use the definition of the derivative to find $f'(x)$ if $f(x)=x^4$
7. Can you guess (or calculate) the derivative $f'(x)$ if $f(x)=x^{10}$
8. Can you guess a rule that would generalize this result so that whenever $f(x)=x^n$, what is $f'(x)$?
9. Use the definition of the derivative to find $f'(x)$ if $f(x)=C g(x)$ for some constant $C$ and differentiable function $g(x)$.
10. Use the definition of the derivative to find $f'(x)$ if $f(x)=g(x)+h(x)$ for two differentiable functions $g(x)$ and $h(x)$.

## Techniques of Differentiation

CONTENT

This material will follow naturally from the material presented earlier if leading questions are asked. It is important to remind students at every available juncture that the function $f(x)$ gives the height at each $x$ whereas the function $f'(x)$ gives something totally different. $f'(x)$ gives the slope of the line tangent to $f(x)$ at each value of $x$. Sometimes if students are not reminded of this, they can get caught up in the analysis of problems and forget what they are doing geometrically.

The so-called “constant rule” and rule for sums and differences should be intuitive; however, the product rule and quotient should be a little more difficult to remember. For this reason it is particularly important that these rules be practiced thoroughly.

The quotient rule in particular requires significant practice in order to keep it memorized. There are two good ways to remember the quotient rule before it becomes second nature. The first is to use a mnemonic. By repeating the following to yourself:

“Low dee high minus high dee low over what’s below (squared)”

it can be recalled that the derivative of a quotient is the bottom times the derivative of the top minus the top times the derivative of the bottom all divided by the denominator squared. The second way to recall the quotient rule simply reduces it to the more intuitive product rule and the yet to be learned chain rule.

PROCESS

This material is best taught through practice, which can be gained in the form of small timed competitions. Students can be divided into two teams and asked to take everything off their desks except for a pencil and blank paper. Then a problem, like “Find the derivative of $f(x)= \frac{5x^2}{x^3-2}$” can be written on the board. The teacher calls on the first hand that is raised, and if the answer is correct then his/her team earns a point. Once students have earned a point for their team, they should be prevented from answering again until the round is over. This will give all of the students an opportunity to participate. Note that rational functions like the one given above or for example $f(x) = \frac{2x^2-1}{x^3+4x-2}$ involve all of the rules studied thus far.

PRODUCTS

Students are best tested on this material by evaluating a large amount of derivatives for rational functions. It is best to continue reiterating what is being geometrically by having students plot the function and its derivative above one-another. For example, students can be asked the following question:

1. Suppose $f(x) = \frac{3x^3}{x^2-1}$. Find $f'(x)$:

a. Answer: $f'(x) = \frac{(x^2-1)9x^2-6x^4}{(x^2-1)^2} = \frac{3x^4-9x^2}{x^4-2x^2+1}$

2. Draw a plot of $f(x)$ and $f'(x)$ on top of one-another:

3. Draw in the tangent lines for the plot of $f(x)$ at the points where $x=2.5, x=1.5, x=0.5$, and $x=0$

4. Write next to each tangent line its approximate slope and compare these with the derivative’s value directly below these points.

In this way students continue connecting the ideas presented here with the geometric nature of derivatives.

## Trigonometric Functions

CONTENT

In order to increase the accessibility of this section, the content should be augmented by adding a discussion of the following key limit:

$\lim_{h \to 0} \frac{\text {Sin}(h)}{h}$

This limit gives the indeterminate form $\frac{0}{0}$ when $h=0$ is plugged in so we would like to use L’Hopitals rule. However, we don’t know what the derivative of the numerator is yet so we will need to think of something else.

The limit above can be found through some considerable trigonometry and analysis, but instead, we make a simple visual argument by looking at the graphs of $\frac{\mathrm {Sin}(h)}{h}$ from $x=0$ to $x=4$:

from this plot it should be pretty clear that we will find:

$\lim_{h \to 0} \frac{\text {Sin}(h)}{h} = 1$

Using this result, and the fact that $\mathrm {Sin} ^ 2 (x)+ \mathrm {Cos} ^ 2 (x)=1$ we can similarly see that:

$\lim_{h \to 0} \frac{1 - \text {Cos}(h)}{h} & = \lim_{h \to 0} \frac{(1 - \text {Cos} (h))(1 + \text {Cos}(h))}{h (1 + \text {Cos}(h))} \\& = \lim_{h \to 0} \frac{(1 - \text {Cos} ^ 2 (h))}{h(1 + \text {Cos}(h))} \\& = \lim_{h \to 0} \frac{- \text {Sin} ^ 2(h)}{h (1 + \text {Cos}(h))} \\& = - \left (\lim_{h \to 0} \frac{\text {Sin}(h)}{h} \right ) \left (\lim_{h \to 0} \frac{\text {Sin}(h)}{1 + \text {Cos}(h)} \right ) \\& = - (1)(0) \\& = 0$

Knowing these two limits is necessary for proving that $\frac{d}{dx} \sin(x) = \cos(x)$ as well as all the other trigonometric derivatives.

PROCESS

Students will understand the cyclic nature of trigonometric functions and their derivatives with practice. In order to make the practice more interesting it is again recommended that students compete to answer simple questions like, if $f(x)= \mathrm {Cos}(x)$, what is $f'''' (x)$? Or for example: Show in two ways that $\frac{d}{dx} (\mathrm {Sin} ^ 2 (x)+ \mathrm {Cos}^2 (x) )=0$.

PRODUCTS

Students can be tested on this and past material in a way that predicts the next lesson. For example, the following are good leading questions:

1. $f(x)=[\mathrm {Sin}(x) ]^1, f'(x)=$?
2. $f(x)=[\mathrm {Sin}(x) ]^2, f'(x)=$? (hint: apply the product rule)
3. $f(x)=[\mathrm {Sin}(x) ]^3, f'(x)=$? (hint: apply the product rule twice)
4. $f(x)=[\mathrm {Sin}(x) ]^{-1}, f'(x)=$? (hint: apply the quotient rule)
5. $f(x)=[\mathrm {Sin}(x) ]^{-2}, f'(x)=$? (hint: apply the product rule and the quotient rule)
6. Can you guess a general rule if $f(x)=[\mathrm {Sin}(x) ]^n$ for finding $f'(x)$?

## The Chain Rule

CONTENT

The chain rule is best explained by the description that derivatives of complicated functions are evaluated “outside to inside.” It is a good idea to give a complicated looking function like:

$f(x)=(\text {Sin}(3x^4+2x) )^3$

The outermost function is the one raising the inside to the third power, so we begin by ignoring the detail inside the parenthesis and write:

$f'(x)=3(\text {Sin}(3x^4+2x) )^2 \frac{d}{dx} \text {Sin}(3x^4+2x)$

This inner derivative again requires the chain rule with the outermost function Sine, so we proceed to break it down:

$f' (x) = 3 (\text {Sin} (3x^4 + 2x))^2 \text {Cos} (3x^4 + 2x) \frac{d}{dx} (3x^4 + 2x)$

Now finally we can simply differentiate this final innermost function term by term using the power rule:

$f'(x)=3(\sin(3x^4+2x) )^2 \text {Cos}(3x^4+2x)(12x^3+2)$

and we’re done!

PROCESS

Teaching the chain rule requires the ability to motivate students to try lots and lots of problems. For example, you could give each student a simple function which requires the chain rule for differentiating like $f(x)=\text {Sin}(x^2)$. Then the student could be asked to plot his/her function and discuss the properties, and then to calculate the derivative $f'(x)$, plot this, and discuss the properties of this as well. Ideally the student will match points of the function with its derivative to confirm that the tangent line looks about right for the slope given by the derivative.

PRODUCTS

To test this material, students should be required to find simple derivatives of composite functions. Again, it is best to continue relating this all back to the basic geometric idea by having students plot the function and the derivative, draw in tangents on the function, and match up points to make sure the slope makes sense.

## Implicit Differentiation

CONTENT

The material here is actually the result of generalizing our conception of the derivative. For example, if we think of the object $\frac{d}{dx}$ as a kind of operator then this all will make much more sense. If one is given an implicit equation like

$yx + \text {Sin}(y) = \frac{y}{x^2+1}$

we can simply think of each side as its own function, the left side we call $L(x,y)$ and the right side we can call $R(x,y)$:

$L(x,y) & = yx + \text {Sin}(y) \\R(x,y) & = \frac{y}{x^2+1}$

These would be a little bit more complicated than we are used to because they are functions of two variables instead of just one. In any case, we know that $L(x,y)=R(x,y)$ and so of course the derivatives should satisfy: $\frac{d}{dx} L(x,y) = \frac{d}{dx} R(x,y)$. Now, to take the derivative of each term we tread carefully:

$\frac{d}{dx} L(x,y) & = x \frac{d}{dx}(y) + y \frac{d}{dx} (x) + \text {Cos}(y) \frac{d}{dx}(y) = x \frac{dy}{dx} + y(1) + \text {Cos}(y) \frac{dy}{dx} \\\frac{d}{dx} R(x,y) & = \frac{(x^2+1) \frac{d}{dx} (y)-y \frac{d}{dx} (x^2+1)}{(x^2+1)^2} = \frac{(x^2+1) \frac{dy}{dx}-y (2x)}{(x^2+1)^2}$

Setting these equal, as prescribed, we obtain that:

$x \frac{dy}{dx} + y + \text {Cos}(y) \frac{dy}{dx} = \frac{(x^2+1) \frac{dy}{dx} - 2xy}{(x^2+1)^2}$

which can then be solved for $\frac{dy}{dx}$ if desired.

PROCESS

One might try to make this material more accessible by relating it to the orbits of planets. For example, draw the following diagram of the “Earth’s Orbit” on the board (note: The actual Earth’s orbit is much closer to a circle than this ellipse):

Then the question could be: If the Sun were to suddenly disappear, what would happen? This will pique the students’ interest and get them thinking a little bit. The idea is that gravity would go away so the earth would no longer be in its orbit. It would instead travel on a straight line tangent to the orbit at the point where it was when the Sun disappeared.

Once students have arrived at the idea that the Earth would fly off along a tangent line, the question should be: What is the equation for the tangent line to an ellipse? Well, the slope of the tangent is $\frac{dy}{dx}$ for the equation of the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ and in order to find this, students should use implicit differentiation. Alternatively, the ellipse can be described as the solution set to the two equations $y = 3\sqrt{\frac{1 -x^2}{16}}$ and $y = -3 \sqrt{\frac{1 - x^2}{16}}$ however an implicit equation is much more elegant. In any event, this description can be used by the students to check their work.

Another fun question might be along the lines of this: Suppose you are riding in a car on the highway with the moon roof open. You put your hand up to feel the air and all of sudden your class ring goes flying off. If you were going around a curve at the time it flew off, which direction would the ring fly? The answer, again, is along the tangent to the curve of your path. And in order to calculate this you should use implicit differentiation.

PRODUCTS

The most straightforward way to test implicit differentiation is by having students find $\frac{dy}{dx}$ for a variety of different implicit functions. In order to keep everything well connected, students should ideally be asked to plot the implicit functions on an $x-y$ grid and find the tangent lines at certain points $(x,y)$ using their determined function for $\frac{dy}{dx}$. This will keep everything nicely connected and prevent students from losing the overall picture of what is going on here.

## Linearization and Newton’s Method

CONTENT

In the text we see that “the tangent line…is a good approximation to the curve” near $x_0$. One logical point of interest is that any other function with the same tangent line is also approximated well by the same line near $x_0$. But what this means is that $f(x)$ is actually approximated well near $x_0$ by any function that shares its tangent there. That means that any function passing through $x_0$ and “traveling in the same direction” as $f(x)$ will approximate $f$ well near $x_0$. A priori there is no reason that we should choose one of these functions instead of another, however the idea of choosing the linear one is that it is simple. A linearization simply means that we are choosing the “simplest” function from among all the functions that pass through $x_0$ in the same direction as $f(x)$.

Newton’s Method is a process for finding where a function becomes zero, by linearizing the function at a point $x_0$ and then solving this linear function for its zero. Then we move to that value which we call $x_1$, and repeat the process to find an $x_2$ and so on. The success or failure of this process depends delicately on the shape of the function at the initial point $x_0$ as well as between $x_0$ and the zero value. For example, if the derivative of the function at any point $x_n$ is ever zero then the linearization there will be horizontal and will not have any zeroes.

PROCESS

The most objective and accessible way to teach linearization is graphically. For example, students could be split into groups of three that are each given a function and a value for $x_0$. For example, one group could be given $f(x)= \mathrm {Sin}(x)$ and $x_0 = \pi$, and another could be given $f(x)=\sqrt{x}$ and $x_0 = 9$. Then the groups should designate tasks to each member: One member will be the artist, one will be the derivator, and one will be the calculator.

The calculator will begin by making a table with a series of points near $x_0$ and the function’s values there, like the example below:

$& x && f(x) = \sqrt{x} && L(x)=f(x_0 )+f'(x_0 )(x-x_0 )=3+ \frac{1}{6}(x-3) \\& 8.7 && 2.95 && 2.94958 \\& 8.8 && 2.96667 && 2.96648 \\& 8.9 && 2.98333 && 2.98329 \\& 9 && 3 && 3 \\& 9.1 && 3.01667 && 3.01662 \\& 9.2 && 3.03333 && 3.03315 \\& 9.3 && 3.05 && 3.04959$

Then the artist will work with this table to draw a careful graph of the function on graph paper near the point $x_0$ while the mathematician determines the equation for the linearization of the group’s function near the point $x_0$. The calculator will calculate and fill in the column of the table for the same $x-$values using the linearization, and the artist will draw in a plot of the linearization (the tangent line). Enough points should be used with enough spacing to be able to see the linearization deviate from the function. This may require a very fine scale, a large spacing, or a great many points. This will be the task of the group as a whole to figure out what window to use and how many points to include.

Newton’s Method can be taught by using it to find zeroes of certain functions. In fact, Newton’s Method can be used to find information like $\sqrt{4.2}$. To find this, we simply search for zeroes of the function $f(x)=x^2-4.2$ beginning at, say, $x_0=2$.

PRODUCTS

Students should demonstrate their understanding of this material by finding the equation for the tangent line of a function and examining how well this approximates the function. This can be done both numerically and graphically, and students should use both techniques.

Newton’s method can be tested by having students search for zeroes of functions beginning with a variety of points. If the students are particularly computer inclined, it is a fun project to have them write a code to carry out Newton’s method. The code would take as an input $f(x),f'(x)$,and $x_0$ and would loop whenever $f'(x) \neq 0$ to set $x = x - \frac{f(x)}{f'(x)}$. At a more basic level, students could simply require the program to loop some predetermined large number of times M. At a more advanced level students could have the program loop until $|f(x)| \le \in$ for some predetermined small number $\in$.

## Date Created:

Feb 23, 2012

Apr 29, 2014
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