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# 5.2: Differentiation

Difficulty Level: At Grade Created by: CK-12

## Tangent Lines and Rates of Change

Problem: The following set of data points give the population, in Millions at a given year.

YearPopulation19001.48195010.59199029.76200033.87200435.89200636.46200736.55200836.76\begin{align*}& \text{Year} && 1900 && 1950 && 1990 && 2000 && 2004 && 2006 && 2007 && 2008 \\ & \text{Population} && 1.48 && 10.59 && 29.76 && 33.87 && 35.89 && 36.46 && 36.55 && 36.76\end{align*}

What was the average rate of change in the 20th\begin{align*}20\mathrm{th}\end{align*} century? What was the rate the population was increasing at the year 2000\begin{align*}2000\end{align*}? Project the population for the state for the year 2020\begin{align*}2020\end{align*} and justify your conclusions.

The focus of this problem is on the decisions made and then writing the justifications for those decisions. The first question is the only one that has a single correct answer. The second question has a couple of options. Students could decide to take the two closest point lower, the 1990\begin{align*}1990\end{align*} data point, and calculate what is actually the average from 1990\begin{align*}1990\end{align*} to 2000\begin{align*}2000\end{align*}. Another, and probably a more justifiable choice, is to use the data point above, as it is closer to 1990\begin{align*}1990\end{align*} and therefore probably more accurate. This is really the key, as students should be working towards an understanding that the closer the points are to each other, the closer the answer will be to the instantaneous rate. Some students may attempt to find an average between the two previous options. As long as students can write a justification for their method, they should be encouraged to find novel solutions.

The last questions leaves the opportunity for the most interpretation. Students should be encouraged to take most of their reasoning from the data given. It may be tempting to bring in other experiences, news items, or personal philosophies, and it is up to the instructor how much to allow, but I would discourage such practices and make the students work strictly from the data.

## The Derivative

There are a variety of standard techniques that are common especially for finding the limits in the form of the definition of the derivative. This may not be exactly the same list as the algorithm for general limits, as the denominator will necessarily go to zero.

• Expand and eliminate. Polynomials will often work with this technique. Example:

\begin{align*}f(x) & =x^3+x+1 \\ f'(x) & =\lim_{h \to 0} \frac{\left ((x+h)^3 + (x+h) + 1 \right ) - (x^3 + x + 1) }{h} \\ & = \lim_{h \to 0} \frac{(x^3 + 3x^2 h + 3xh^2 + h^3 + x + h+ 1) - (x^3 + x + 1)}{h}\end{align*}

Then distributing the negative and re-ordering for clear cancelations:

\begin{align*}\lim_{h \to 0} \frac{(x^3 - x^3 + 3x^2h + 3xh^2 + h^3 + x - x + h + 1 - 1)}{h} = \lim_{h \to 0}\frac{(3x^2h + 3xh^2 + h^3 + h)}{h}\end{align*}

Then cancel out a factor of \begin{align*}h\end{align*} and evaluate the limit:

\begin{align*}\lim_{x \to 0}3x^2+3xh+h^2+1=3x^2+1\end{align*}

Something to caution students about is consistent use of parenthesis. This is especially critical in making sure the negative gets istributed correctly to cause the proper cancellations.

• Multiply by the conjugate of the numerator. Usually used for radicals. Example:

\begin{align*}f(x) & =2-\sqrt{x-5}\\ f'(x)& = \lim_{h \to 0} \frac{2 - \sqrt{(x+h)-5)} - \left (2 - \sqrt{x-5} \right )}{h} \\ & = \lim_{h \to 0} \frac{\left (- \sqrt{(x+5) - 5} \right ) + \sqrt{x-5}}{h} \times \frac{- \sqrt{(x+h) -5} - \sqrt{x-5}}{- \sqrt{(x+h) - 5} - \sqrt{x-5}} \\ & = \lim_{h \to 0} \frac{(x+h -5) - (x-5)}{h\left (\left (- \sqrt{x-h-5}\right ) - \sqrt{x-5}\right )} \\ & = \lim_{h \to 0} \frac{h}{h\left (\left (-\sqrt{x-h-5}\right )-\sqrt{x-5}\right )}\\ & = \lim_{h \to 0} \frac{1}{- \sqrt{x-h-5} - \sqrt{x-5}} \\ & = \lim_{h \to 0}\frac{1}{- \sqrt{x-5} - \sqrt{x-5}} \\ & = \frac{-1}{2 \sqrt{x-5}}\end{align*}

Again, keeping a close watch on the negatives is key to getting a correct answer.

• Use identities and rules of trig functions, logarithms and other functions. Example:

\begin{align*}f(x) & = \cos (3x) \\ f'(x) & = \lim_{h \to 0} \frac{\cos(3x+3h) - \cos(3x)}{h} \\ & = \lim_{h \to 0} \frac{\cos (3x) \cos (3h) - \sin (3x) \sin (3h) - \cos (3x)}{h} \\ & = \lim_{h \to 0} \cos (3x) \frac{\cos (3h) - 1}{h} - \lim_{h \to 0} \sin(3x) \frac{\sin(3h)}{h} \\ & = \lim_{h \to 0} 3 \cos(3x) \frac{\cos(3h) - 1}{3h} - \lim_{h \to 0} 3 \sin(3x) \frac{\sin(3h)}{3h} \\ & = 3 \cos (3x)(0) - 3 \sin (3x)(1) \\ & = - \sin(3x)\end{align*}

Recalling a trig identity and a couple of limits from a previous chapter.

## Techniques of Differentiation

There are three rules presented in this chapter are some of the most used throughout a first year calculus course. It is important then to get lots of practice with selecting and using each tool. Most are straightforward to implement, but students seem to have the most difficulty with the quotient rule. A couple of hints for the quotient rule:

• Remember subtraction is not communicative. While the product rule can be used with the terms in any order, the quotient rule must always be used the same way.
• Be consistent and thorough with parenthesis. Common errors include the incorrect distribution of the negative stemming from not being clear with groupings.
• Don't forget about doing algebra correctly. It's easy for students to get very involved with applying the power rule inside of the quotient rule and focusing completely on the tools they are learning, and then incorrectly square the denominator.
• Don't use it. While sometimes the process requires the use of the chain rule, in a future section, students tend to make fewer mistakes if they can simplify the function in advance, or use a negative exponent to make the use of the product rule possible.

## Derivatives of Trigonometric Functions

Here is a completely contrived problem, but very nicely illustrates the correlation for the trig functions.

Problem: A car is on a circular track with a radius of 1km maintaining a perfectly steady speed of \begin{align*}1km/h\end{align*}.

1. Plot two graphs. The first showing the vertical displacement away from the center and the second showing the horizontal displacement from the center, both as a function of time.
2. Plot two graphs, this time showing the vertical and horizontal velocity. Hint: you may find it helpful to first plot the “easy” points, the ones on the axis, and the ones half way along each quadrant. The Pythagorean theorem may come in handy here.
3. Make a prediction about the graph of the vertical and horizontal acceleration against time and the direction of the acceleration of the car.

This is a challenging question. There are a couple of ways to plot the first two graphs. One is to use right triangles and the trig functions. This is a little bit circular, of course, as the graphs are going to be the graphs of sine and cosine respectively. Another, more intuitive, way to do it is to use the endpoints and the \begin{align*}45^\circ\end{align*} and \begin{align*}30-60-90\end{align*} right triangles. That gives \begin{align*}16\end{align*} points and should result in enough information to make a curve. It's ok if the students begin to graph, recognize the function and complete the graph from there. The same procedures apply for the second question. The big key here is to have the students recognize another relationship between the trig functions, not only that sine and cosine represent the coordinates around the unit circle, but also that there is a natural way to relate the derivatives of each function. Another benefit, although it may also create a challenge, is having the students work with a situation where the speed is constant, but the velocity, being a vectored quantity, is changing.

After working with the velocity in this manner, some students may come to the proper conclusion about the acceleration, but if not it is not a problem. This is more of the process of “stretching the mind” and giving students a problem that is maybe one step further then they are comfortable with and asking them to give their best prediction and justification. The acceleration vector for the car is always going to be constant, and pointing in towards the center of the circle. This makes the \begin{align*}x\end{align*} and \begin{align*}y\end{align*} component vectors the legs of the right triangle the acceleration vector creates. Another fun question: there is a helium balloon tied down and floating freely in the car. Which way is the balloon leaning as the car turns left around the circular track? Answer: the balloon leans to the left. The air pressure will be greater on the right hand side of the car due to centrifugal forces pushing the balloon to the area of lesser air pressure on the left.

## The Chain Rule

Now that students have most of the tools for differentiation they will use, it's time to look at putting many of those techniques together. This can be a daunting task for some students as it not only means recognizing which tool to use, but what order, and with no directive on how many times they may need to use it. Let's look at a rather complicated problem as an illustration.

\begin{align*}f(x) = \frac{\sqrt{3x \cos (x) - \text{ln}(5x)}}{\sin^2 (4x^3)}\end{align*}

To find the derivative of this function multiple applications of the chain rule, product rule and quotient rule. A couple of problem solving hints:

• Work from the “outside in.” Meaning that the grouping functions that are outside get treated before the functions that are inside.
• For nested rules, let the rule dictate which rule you need to use next. Don't immediately go to making a list of all techniques needed. If you are in the middle of using the quotient rule, and you need a derivative of the top function, then look at what is needed to take the derivative of the top function.
• Don't lose your place. Because a rule may get started, and then not finished until after a number of other rules are applied, don't lose track of where you are in that rule. Something that may help is labeling the separate parts, writing their individual derivatives and then putting it all together in the end.

I'll solve this problem showing the maximum amount of work for clarity.

First, since the fraction bar extends the whole way, the quotient rule needs to be applied first.

\begin{align*}\frac{p'(x)q(x) - p(x)q'(x)}{(q(x))^2} , p(x) = \sqrt{3x \cos (x) - \text{ln}(5x)}\end{align*} and \begin{align*}q(x) = \sin^2(4x^3)\end{align*}

Now we need to take the derivative of each function individually. Since this is going to be involved, lets look at \begin{align*}p(x)\end{align*} first. It is useful to re-write the function with a fractional exponent and apply the chain rule:

\begin{align*}p(x)=(r(x))^{\frac{1}{2}}, p'(x)= \frac{1}{2}(r(x))^{\frac{-1}{2r'}}(x)\end{align*} where \begin{align*}r(x)=3x \cos(x)- \text{ln}(5x)\end{align*}

As the chain rule states, we then need the derivative of the inside function, \begin{align*}r(x)\end{align*}, but it is important to recognize that the first term is a product of two functions and the second term is a composition of functions. So applying the product rule for the first term and the chain rule for the second results in: \begin{align*}r(x) & =m(x)n(x)-c(d(x)), m(x)=3x n(x)=\cos(x) c(x)= \text{ln}(x) d(x)=5x \\ r'(x) & =m'(x)n(x)+m(x) n'(x)-c'(d(x)) d'(x), m'(x)=3 n'(x)= -\sin(x) \\ c'(x) & = \frac{1}{x} d'(x)=5 \rightarrow r'(x)=3\cos(x)-3x \sin(x)- \frac{1}{5x} \times 5\end{align*}

This is the end of the line for the top, now its time to look at \begin{align*}q(x)\end{align*}.

\begin{align*}q(x) & =(\sin(4x^3))^2, q(x)=g(h(j(x))), \\ g(x) & =x^2 h(x)=\sin(x) j(x)=4x^3 \end{align*}

It may not be clear on first inspection that this is actually a composition of three functions together. It is this reason why it is often useful to re-write exponents for trig functions “outside” using parenthesis. Writing it out using the chain rule with function notation:

\begin{align*}q'(x) & =g'(h(j)) h'(j(x)) j'(x),g'(x)=2x h'(x)=\cos(x) j(x)=12x^2 \\ q'(x) & =2 \sin(4x^3)\cos(4x^3) 12x^2=24x^2 \sin(4x^3) \cos(4x^3)\end{align*}

Now it's time to put all of it together. Since there are no more derivatives left to take we can work from the bottom and fill in the derivatives that are called for in function form.

\begin{align*}f'(x) = \frac{\left ( \frac{1}{2} (3 \cos(x) - \text{ln}(5x))^{\frac{-1}{2}} \left (3 \cos(x) - 3x \sin(x) - \frac{1}{x} \right ) \right ) \sin^2 (4x^3) - \sqrt{3x \cos(x) - \text{ln}(5x)} 24x^2 \sin(4x^3) \cos(4x^3)}{\sin^4(4x^3)}\end{align*}

This was primarily an illustrative practice. I do not recommend such an involved problem, except maybe as a bonus or a special problem, but it does expose either bad habits with showing and tracking work, as well as the importance of continually letting the rule in use dictate the next step.

## Implicit Differentiation

Implicit differentiation is really a fancy instance of the chain rule. The key to having success solving more challenging problems that are written implicitly is to follow the clues and processes set up in previous sections, only remembering the derivative terms that need to be chained at the end. An example:

ex. \begin{align*}y=x^4 y^3+x^3 y^4\end{align*}

Clearly this equation can't be solved explicitly, which is always a good thing to check. Now taking the derivative implicitly is going to require the use of the product rule for the two terms on the right hand side. Each of these then requires the use of the chain rule as part of the implicit differentiation.

\begin{align*}\frac{dy}{dx} = \left (4x^3 \frac{dx}{dx} y^3+x^3 3y^2 \frac{dy}{dx} \right ) + \left (3x^2 \frac{dx}{dx} y^4+x^3 4y^3 \frac{dy}{dx} \right )\end{align*}

Note that I included the \begin{align*}\frac{dx}{dx}\end{align*} derivative term. I encourage students to do so, and then cancel it out later. This provides consistency with applying the chain rule, and avoids the trouble that can happen when students do not know when they need to “do it” and when they can “ignore it.” The next step is to cancel out the derivatives that are equal to one, and then group the terms so we can prepare to factor the derivative term of \begin{align*}y\end{align*} with respect to \begin{align*}x\end{align*}.

\begin{align*}\frac{dy}{dx} = 4x^3 y^3 + 3x^2 y^4 + 3x^3 y^2 \frac{dy}{dx} + 4x^3 y^3 \frac{dy}{dx}\end{align*}

Subtracting to get the derivative term on the same side, then factoring:

\begin{align*}\frac{dy}{dx}-3x^3 y^2 \frac{dy}{dx}-4x^3 y^3 \frac{dy}{dx} & =4x^3 y^3+3x^2 y^4 \\ \frac{dy}{dx} (1-3x^3 y^2-4x^3 y^3 ) & =4x^3 y^3+3x^2 y^4 \\ \frac{dy}{dx} & = \frac{4x^3 y^3+3x^2 y^4}{1-3x^3 y^2-4x^3 y^3}\end{align*}

It is useful to think of the derivative terms as quasi-variables. They can be added, multiplied and factored just like variables. Having this understanding will help with separable differential equations later on.

## Linearization and Newton's Method

The topics of linear approximations for curves, and then the use of such linear equations to approximate solutions for difficult equations may be a tough sell in today's world. The topics keep on reappearing on standards lists, and occasionally show up on the AP examination, and this keeps the flame going for these topics. Students, having grown up in an era with computers and graphing calculators, all with symbolic solvers, often ask “Why?” With the expectation from the exam writers being that students know it, the answer becomes “Because.” But this does present an interesting question: If these methods have been made somewhat obsolete by technology, how do you test mastery?”

With so much of a high school calculus class being driven by the AP examination, it is useful as a problem solving skill to predict what types of questions can be asked in a reasonable manner. The format of the test does restrict the type of question heavily, and therefore keeping in mind what types of questions can be asked may prove helpful. It is not reasonable to expect students to be able to have mastery of all types of questions, in all situations, in calculus in only a year. As the focus of a class changes, for instance a high school AP class, a university level year one class for social science and biology majors, and a year one class for math, physics and engineering students, one can see how the longer format questions change.

Specifically for this section, how do obsolete questions get asked? There are two major ways for these questions to show up. First is to require exact answers with irrational numbers. Since even in the calculator legal sections the technology is restricted to calculators without symbolic solving systems, requiring answers in exact form is a way to enforce hand-working of the problems. The second method is to put the problems in a calculator illegal section of the test. This places another set of restrictions, as the expectations of what the students can be asked to do changes when no calculator is allowed.

Therefore, it is valuable to work on problems, especially in this section, with a variety of calculators allowed, and formats required for answers. Students should be asked to solve problems with use of a graphing calculator, and without. Also, since many university math and science departments are not allowing graphing calculators in their lower division classes, but are requiring a scientific calculator, it may be useful to also practice using a scientific function calculator. It is also helpful to require students to work problems with exact irrational numbers throughout problems of various kinds, getting used to the sometimes a variable, sometimes a number treatment of such elements.

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