It seems that most calculus tests I receive to check my programs with, and with regards to graphing a function by hand, they always have you find: ” critical points”. Then the answer is always just what “x” equals. You factor the first derivative f ‘(x), find the value of x or x’s and mark it down on your test problem to get it correct x=2 or x=5 or whatever. If I was your professor (you wish) you’d have gotten only partial credit because you found the “critical numbers” and not the “critical points”. However, critical points are actually the value of (x,y). You find the values of x, from the first derivative, plug those values into the original function f(x) to find the value of y, and you have the critical point or points (x,y). When tests ask for critical numbers the professor actually means critical numbers. There is a difference!! In my program I find both for you critical numbers and critical points (step by step of course) and leave it up to you as to what or how your professor teaches this, in most cases teaching it incorrectly, what else is new?
Definition of Derivative, Calculus App, TI89 Titanium
Raw Transcript
This is a video on the definition of the derivative also called a quotient, difference quotient. A very difficult program wouldn’t be difficult for you and me, if you have all day, all night and several days to study it and get the handle on it and really think about it). But this program may help in your calculus endeavors. Gonna get started, press 2nd Alpha (this shows that you can enter letters in the calculator). My code for getting into my program is index then you press alpha again to get number 8 and the (). Should be index8 (). Press enter and that takes you to the menu, where you can scroll down and select what you want. We’re working with definition of derivative here, select and press enter and put in formula or function. Press Alpha first then enter everything in the boxes on the program. We’re gonna do 3*x^2x+1 go to ok or change it if you want. Write down the formula that comes up. First thing to do when writing a problem is to write the formula down and then you start substituting x+h for every x in the function, which I’ve done here. 3*”x+h” or parentheses actually, squared, minus “x+h”+1 and then the original formula all over h.
F’ (x) =lim
h>0
=6*x+3*h1
@h=0
Answer
=6*x1
Everystepcalculus.com, enjoy my program visit my website.
Difference Quotient, Calculus App, TI89 Titanium
Difference Quotient, Calculus App, TI89 Titanium
Raw Transcript
Hello, this is Tom from everystepcalculus.com. I’m gonna do the difference quotient for you today or what others call the definition of a derivative.
We’re gonna get started. Turn calculator on, press 2nd Alpha and put index the alpha again, then 8, then () and press enter. Up comes my program menu, you can see the things I’ve programmed so far. There is a graphing section with the calculus and graphing, maximum and mins among others. But today we’ll be doing a definition of a derivative or difference quotient, which ever you know, either one will take you to the program. Press enter and you are in the actual program. Every time you see pause here, you can enter. We’re gonna put in an example, for instance, (you have to press alpha first when you put anything for these boxes) I like to use 7*^27*x. Now you’re gonna substitute for every x in the function, x+h
Should look like this: f(x) =7*x^27*x select ok or change it if you want. This is the formula, the f ‘ of x = the limit as h goes to 0 h represents the change of x.
Formula should be: f’ (x) = lim
h>0
= f (x+h) f (x)
h
h represents x
Jot down on your paper then put in your function. Notice I’ve substituted “x+h” for every function and the original function all divided by h.
F’ (x)
7*”x+h”^27*”x+h”
= (7*x^27*x)
h
Very complicated program, took me a long time, maybe a month to do this program. Worked a lot with research to do it, calculus books and such. So, you’re gonna write this on your paper, but you’ve taken the x squared of the function and worked it out.
F’ (x)
7*x^2+14*h*x+7h^2
““ (7*(x+h))
(7*x^27*x)
h
Keep writing these down and eventually you’ll get:
F’ (x) = lim
h>0
=14*x+7*h7
@ h=0
Answer
= 14*x7
And you’ve gotten a hundred percent on that problem and you look like you know what you’re doing and everything else.
Everystepcalculus.com you can go to a new problem, main menu or you can quit or do what you want.
Enjoy my programs
Curl Program
Raw Transcript
Graphing: Maximum & Minimum Program
Video Example: Critical Points for Graphing Program
Raw Transcript
So, this video is on critical points and critical numbers, with regard to graphing by hand in calculus. Calculus one, or all through calculus I guess. Ahh Let’s get started. You press second alpha on your calculator titanium. See this box here which turns black which shows you that you can enter letters. And the code for my menu of course is I n d e x, and then you have to press alpha to get back to numbers and parenthesis, and you’re into my menu. My menu has many many things in it. Right now we’re talking about critical points, so – were gonna go – uhm – to critical points. Pull up the program, and I tell you to mark on your paper right away, on your test paper or whatever, ya know, a graph with these, ahh all labeled in numbers so you can mark down whatever comes up in my program, and then you can, ya know, connect the dots and have your graph completed – by hand. So we’re going to enter a function, a from a test, we’re going to have to – for any of these boxes that come up in my programs you have to add alpha – you have to press alpha first – so we press alpha, and we’re going to put in x cubed minus six times x squared, plus nine times x and plus two, and then press enter twice – and I show you what you’ve entered, in case you want, made a mistake, you can go back and do it, you press enter again, you can, ok or change it – I’m saying its ok – my programs also, you can press the number before these what the choice is, you can scroll down, or and then press enter, but you can also just press the number and it will go right to there. So were at critical points, we want critical points in the menu, instead of scrolling we’re going to just press three on the calculator – and – and I – discuss a critical points and numbers in a blog in my web site, so check that out. Most tests, come up with, ask you for critical points, and there’s a difference between critical numbers, critical points – and I discuss that – here I put a little bit of information about it. Critical point is really an xy point on the graph, and critical numbers are – are on the x axis, just what the x value is – however – professors and tests I’ve seen – uhm – ah – you know use both and – and it’s not correct, their different. Ahh, So – in critical numbers, your gonna – you’re going to set the uhm, first derivative to zero, and then solve for x – so we factor the first derivative, here, and we come up with the critical numbers – x equals three or x equals one – that’s what you’d put down – You put everything on your test, just like this. You can’t find the first derivative unless you put the function down – then you find the first derivative – and go from there – uhm – I do the – I find critical numbers and critical points on my programs – so – we add – we take three – one of the critical numbers of three – and plug it into the first function – three cubed, times, minus – six times three squared, plus nine times three, and you write this on your paper and you come with two – y equals two – so the first critical point is three and two – second one you plug in one for critical number into the primary function, and come up with six – so the second critical point is one and six – and then, it takes you back – you can find more parameters here, press one, you can get more parameters, and go whatever you want – local maximum min – Intervals of increase decrease – inflection point – what ever you need to complete the graph
Equation of Tangent Line to a Curve Program
Raw Transcript
This video is gonna be on the equation of a tangent to the x point on a curve. This is the line when you have found the first derivative and you find the slope. This is the equation that is in the line at the point where you find the slope. This program I pretty neat, check it out on my website. Turn calculator on and clear screen to get to the menu of the program by pressing F1 8. Now add index 8 () to get to the menu of the program. Press 2nd Alpha, enter index, press alpha again to get back numbers and parentheses add 8 (). Here we are, select the equation of a tangent line, scroll down and press enter. It shows you what programs are added here. Enter the function 8*x to the 3rd power + 6*x+9
You can change or press ok if you want. Now you want a point, maybe at point 3, press Alpha 3 (it shows x=3). If youmade a mistake you can change it or say ok.
Function: F (x) = 8*x^3+6*x+9
F (x) = slope
= 24*x^2+6
The derivative o 6x is 6 f’ (x) =24*x^2+6
@ x= 3
F’ (3) = 24*”3”^2+6
=222
You have to go back and put 3 in the function to find y
Y = f (x) = 5s^2+6x+9
@ x = 3
Y = 8*’3”^3+6*”3”+9
=243
So (x,y) = (3,243)
Y=243 (222 is the slope)
243 = mx+b
243= 222(3) +b
B=243666
=423
Y = 222x+423
Video Example: Equation of a line program
Raw Transcript
Ah, this video is going to be on the equation of a line, with regards to my fabulous programs that I’ve programmed on the TI89 calculator, and ah, these are pretty simple calculations, but it’s easy to forget how to do it. It’s nice to have a program to be able to do it. So, I’m going to clear the calculator here and we’re going to put in; 2nd Alpha, 2nd has to appear here, and then the alpha has to, to become darkened to go for the letters. And I have to put that in the entry line – I n d e x – and you will to if you buy the programs, ah alpha, it goes back to the number system, and then put the closed parentheses in, and press enter and we’re into the ah menu. You’ll notice there are many, many things on this menu. Definite integral, ahh derivatives, you know in algebra all of derivatives, ah transforming ah problems. Curl, product rule, quotient rule, whatever, but we’re going to do equation of a tangent line, you can press alpha e, or you can scroll down like this, and ahh, and press enter equation of a tangent line. Ah we’re going to enter the function, five times x squared, plus six times x, plus one let’s say – notice I forgot to press alpha, first, which is easy to do on this calculator, and so I’m gonna go back with two, I’m going to change it, and I’ll press alpha, five time x squared, plus six times x, plus one, and that’s better, five x squared plus six x plus one that’s the function – now say it’s ok – so we’re gonna, I generally press the one before the choice – and then we’re going to enter the x y component, you know, x, the values for x and y, so let’s do that now, let’s do it. You have to the parenthesis in first – you have to go alpha again – and put, let’s go three comma sixtyfour, something – close out the parenthesis – and let’s see – now we got three, x y z equals three sixty four, that’s cool – press one – and we have the original function – we find the derivative of it, the slope – which equals ten x plus six – at three – our x choice that we had, is the derivative, derivative at three is ten, and then add the three to the x and there’s the – thirty six is the slope which is m – press enter again – shows you the formula – you write this stuff on your paper as you go – and at y equals sixty four – which was our choice – sixty four equals m x plus b, so sixty four equals thirty six that was the slope – times 3 – which was x – plus b, sixty four minus one o eight minus forty four – here’s the formula – y equals 36 s+ minus 44 – for that point pretty neat, huh? everystepcalculus.com – check it out
Natural Log Program
Raw Transcript
Quotient Rule Step by Step
Raw Transcript
Product Rule Step by Step
Raw Transcript
so i’m going to demonstrate the uh… product rule on the titanium written you know with regard to my program start program a calculator uh… and to we need to get to the home screen here and you type index that i’m gonna clear this out and reapply it so they can show you how to do that you go second alpha get that little dark mark in there to show you you’re gonna put letters in the calculator i_n_d_e x and then alpha again to switch the numbers in parentheses and your into my programs there’s a menu of many many things depending on what i want to put in there but you know product rule like this one chain rule, quotient rule, quotient um difference quotient, limits trig integrals, derivatives log of base a or you know natural log derivatives, derivatives of thosebut anyways, we’re going to go team also goes straight up and then you can scroll down to go here’s velocity and stuff that you would need in calculus, um a lot of it were going to do the quotient rule in the next video but anyways were going to do the product rule now you can see that I’ve highlighted that, while that’s loading the formula for the product rule is of course h of x because you’re doing two functions f of x and g of x h prime of x is this uh… formula here prime of x g event epa vexed times g prime objectives the part of it for the and then were going to enter in the parentheses like this country an example so we have to go alpha and then enter the first parentheses and then you can enter whatever let’s do 5 times x squared plus six closed parentheses without the prince of sleaze which shows you what you’d entered the program you write that down in your paper dash if you think you want it if it’s okay if you like if you made a mistake and go back and change it whatever so saying it’s ok so each prime although it had to be good at wildness times the either function plus the other function times riveted the other functions of that so you write that on your paper ten x etcetera etcetera we come to the three h primal inferences that explicitly plus the derivatives at this when you add it up or multiply it up at fortyfive expert ninety extra eighteen and that’s the product rule opt regarding my programs check them out of my website at least of calculus dot com
Chain Rule Program Step by Step
Raw Transcript
Equation of Tangent Line on TI89
Equation of Tangent Line to a Curve
Raw Transcript
This video is gonna be on the equation of a tangent to the x point on a curve. This is the line when you have found the first derivative and you find the slope. This is the equation that is in the line at the point where you find the slope. This program I pretty neat, check it out on my website.
Turn calculator on and clear screen to get to the menu of the program by pressing F1 8. Now add index 8 () to get to the menu of the program. Press 2nd Alpha, enter index, press alpha again to get back numbers and parentheses add 8 (). Here we are, select the equation of a tangent line, scroll down and press enter. It shows you what programs are added here.
Enter the function 8*x to the 3rd power + 6*x+9
You can change or press ok if you want. Now you want a point, maybe at point 3, press Alpha 3 (it shows x=3). If you made a mistake you can change it or say ok.
Function: F (x) = 8*x^3+6*x+9
F (x) = slope
= 24*x^2+6
The derivative o 6x is 6 f’ (x) =24*x^2+6
@ x= 3
F’ (3) = 24*”3”^2+6
=222
You have to go back and put 3 in the function to find y
Y = f (x) = 5s^2+6x+9
@ x = 3
Y = 8*’3”^3+6*”3”+9
=243
So (x,y) = (3,243)
Y=243 (222 is the slope)
243 = mx+b
243= 222(3) +b
B=243666
=423
Y = 222x+423
How Professors teach compared to my programs
Here is what you get as an answer usually when you ask a question on Calculus or even Physics in my experience and evidently the asker was satisfied. The person answering is a professor at a college. This girl asked on line for help on what the derivative of 4cos(5x2) was. This is a chain rule problem. How would you like it answered?

Best answer as selected by question asker.
For a function f(x) = g(h(x)), express h(x) as y.
Then f(x) = g(y), f’(x) = [d {g(y)}/ dy]*(dy/dx).
Here we have to find the derivative of f(x)= 4 cos (5x2).
Let y=5x2, this gives f(x)= 4 cos y
f’(x)= [d (4 cos y)/dy]*[d(5x2)/dx]
We also know that the derivative of cos x= sin x.
=> [d (4 cos y)/dy]= 4 sin y
[d(5x2)/dx]= 5
Therefore f’(x)= [d (4 cos y)/dy]*[d(5x2)/dx]
= (4 sin y)*5
=4*sin (5x2)*5
=20 sin (5x2)
Therefore the derivative of 4 cos (5x2) is 20 sin (5x2)
Calculus
I had a person question me over the difference quotient. He couldn’t load 4x3 into my programs. I had not programmed that “straight line” into my programs I had only programmed something with x^2 (curve). He gave me a you tube video feed that 4x1 was a valid function. I guess it is after watching the feed. So I programmed it for him and now you. I’m a guy who thinks practical about things. I programmed my calculator because I was smart and allowed to use the calculator for tests.
Point Slope Form: The relation to calculus
The equation of a line to a point on a curve (point slope form) includes the slope and the position of that line on that curve function. It’s better than the derivative because the derivative only tells us the slope. Again in Algebra the professor forgot to tell us the importance of that and the relationship to the derivative. Didn’t make it interesting enough to sink in and how it relates to the real world.
You have a function. Has to have x^2 in it to be a curve from my understanding, Example y or f(x) = 3x^2
Graph that and you have some form of curve in this case a “valley” parabola, (my own word), 3x^2 and you have a “mountain” parabola (again my own word).
Pick any point “(x,y)” Example: (3,12)
Point = (3,12)
x = 3
y = 12
Find the derivative: f(x) = 3x^2
f’(x) = (2)(3)x^(21)
= 6x^(1)
= 6x
Compute the derivative at the point “x”
f’(3) = 6(3)
= 18 = m = slope
Point slope form = y = mx + b
y = 12 so:
12 = mx + b
m = 18
12 = 18x + b
x = 3 so:
12 = 18(3) + b
= 54 + b
b = 12 – 54
= – 42
y = mx + b
= 18x + 42
If you graph this equation along with the original function you’ll see the tangent line to that point on the curve
The slope = 18/1 (rise over run)
The angle of that tangent line = tan^(1)(18/1) = 86.8 degrees
(make sure your calculator mode is in APPROXIMATE and DEGREES)
Fabulous and exciting, right? lol Tom
p.s. You’ll love my programs
Have a test or quiz on point slope form? Here is a video example using the programs on the TI89 calculator: (Click Links below)
What is a derivative?
Ask anybody “what is a derivative?” and you’ll quickly find out that nobody can tell you exactly what it is in no uncertain terms without any question. If fact, most people you ask who have taken calculus can’t even come close. Think I’m wrong, try it and you’ll find out what I did, it took me eight years after college to find out and even then not exactly. That’s pathetic and unacceptable in my opinion. Of course I was programming the TI Calculators in calculus at the time so I had some interest to even ask the question. I mean, who cares right? No one out of college will ever calculate a derivative or integral again in any job outside of reteaching it as a professor, so I/we understand that.
I had some interest because of programming step by step calculus into the calculators and while teaching tennis to this guy named Mike, I find out – he at one time worked at NASA. At the time of me teaching him tennis he had left NASA was a professional black jack player. Went all over the world making money at black jack in the casinos. I was also fooling around with on line poker at the time and asked him how come black jack and not poker? He said poker was too much gambling and black jack is relatively sure. He said he used differential equations to help him count cards and change the odds in his favor. Anyway I asked him what was a derivative? He said immediately that it was the slope of a tangent line to a curve. I said “but when I graph the derivative there is a line – but no tangent line and the slope is off.” He said the graph of the tangent line is meaningless, “of no value”. He said when you compute the derivative of a function at an “x” value you come up with a number and that is the slope of the tangent line at that point on the function. I thought even that was fabulous and an eye opener, but we had finished picking up tennis balls and so I let it go and started to again teach him tennis.
After that moment, I kept thinking and thinking and thinking of what he said and then it dawned on me. The number you get when computing the derivative at the chosen point “x”, no matter how deep the original function is the numerator of a fraction with the denominator = to 1. That was rise over run. If the numerator is 12, that is 12/1. You go 12 spaces up the “y” axis and one space over on the “x” axis. Draw a line from that point through the origin of a graph (0,0) and that is the exact angle or slope of the line of the derivative. For that line to me tangent, it must touch the curve at only one minute point, so that line has to be transposed to do that, however it will still be the exact slope. To transpose that line you have to compute the equation of that line to the desired point (point slope form) and then graph that function and you have the perfect picture of what a derivative is. That’s fabulous. (Equation of a line at a point on a curve is in my calc1 programs). Now at the next party you know all there is of “what’s a derivative” and can look like a nurd, I mean nerd. lol, Tom
p.s. (The exact angle of that slope by the way is tan^(1)(12,1), Fabulous!!
First day of calculus I
In my first day of calculus I – the chalk was flying. That professor started with the fundamental theorem of calculus, probably said “if it exists” a hundred times, and never let up from then on. It got worse from then on and never better. I, with others – I’m sure – sat there in disbelief and in a fog. I was 50 years old that day. I remember 3 semesters later I turn to the guy next to me and ask, “What the hell is a derivative?” He whispers, “I think it’s an angle of a line or something”.
Try it yourself go up to any one of your friends, ask them first if they’ve taken calculus and then ask them what a derivative is, and see what the say or don’t say. That’s the way all of my professors taught in my experience of college, I never enjoyed one class. If it was me – and my inability to learn in a class – that kid would have told me what a derivative was without any question as well as others in the class, however nobody knew (I asked several others after that kid) they didn’t know either. We were all three semesters into calculus and didn’t know what a derivative was. What the hell is that? Isn’t knowing what a derivative is, in no uncertain terms, more important than the fundamental theorem of calculus???
Professors
In my experience professors teach theory and proofs and then test on mechanics which somehow (the mechanics) get lost in their lectures. Also, maybe understandably, because of tenure and boredom in their jobs they might tend to get away from – or not get down to – the struggling students level when they teach. In my programming of a problem, I can’t skip steps to the solution, somehow every step must be addressed and accomplished. That’s what make my programs as designed so essential, helpful and perfect for the student. Enjoy my programs, Tom
Calculus and you
The difference in my programs as created for myself compared to other programs on the TI site or internet itself, is that the emphasis is on the mechanics, the perfect system, step by step, to a problem's solution. No theory, no proofs just be able to do the problem for my tests or homework and move on. If after that I wanted to know more about a problem, the whys and wheres etc. fine, but at least I could do the problem. You will learn with my programs as hundreds have for tests or homework but with no emphasis on theory, or proofs. In my experience in college there was no time for true learning or understanding, just cram, test and forget.I came to the unwanted experience at the age of 50 going to college for the first time, that college was teaching me everything I never wanted to know about things that made no difference. Like drinking from a fire hose. Too much material and no time. Enjoy my programs, Tom