Archives for 2014
Limit to Infinity Example 2-Solved by TI-89 Video
Lim (7x^3+9(x^2+3)
–> Infinity
Raw Transcript
Hello Everyone, Tom from everystepcalculus.com and everystepphysics.com. I’m going to do another Limit to Infinity problem. Index 88 to my menu. Choose Limits. Scroll to compute limit. Enter the function by pressing alpha. This test problem is seven times x cubed plus nine. Uh oh, I forgot the left parentheses. Press second, go there quick. Divided by x squared plus 3. Add the Infinity indication. Choose the other button, and the yellow register is underneath. See the yellow letters, that’s what that does for us to that register when you press the yellow button there. I always show you what you’ve entered, you can change it if you want. which you better you can change it if want. You’re dividing everything, every term in that problem by the highest order term in the denominator. This here equals this and then we add infinity to these, it’s going to come seven because anytime you take a number. So when you multiply times Infinity it becomes Infinity. So here’s Infinity and second one is zero.Numerator Limit is Infinity. Denominator divided by the highest order term of denominator is this. That equals. When you’re add Infinity, that equals one zero. The Denominator limit is one. The Problem limit is Infinity then. Infinity over one. Pretty neat, huh? everystepcalculus.com, visit my site and buy my programs. Thank you
Limits to Infinity on TI-89 Q4
Lim (7x^3+9)/(x^2+3)
Raw Transcript
Hello Everyone, Tom from everystepcalculus.com and everystepphysics.com. I’m going to show you how to do a Limit at Infinity, on this video. I’m doing several of them so you can get a great idea of what happens.Index 8 to get to my menu. Limits, Choose Compute Limit. Enter the function. Alpha Seven times x cubed plus nine. Oops forgot the other parentheses,
So we press second and the arrow here to get quickly over to the beginning. Add our parentheses and press second and the right arrow to get back where we were. precedent for a second and the right. Divided by parentheses x squared plus three. Need to add the Infinity. Yellow button here.I show you what you’ve entered. You can change it if you want. I say it’s okay. Again, we divide all the terms by the lowest ordered by the lowest quartered term in the denominator. And that equals this for the numerator. And when you do Infinity, the Numerator Limit is Infinity.Denominator…you take those terms and divided by x squared also. Equals one. Infinity divided by one is Infinity.That’s the answer. Problem Limit.
Pretty neat, huh? Visit my site and buy my programs. Thank you
Limits at Infinity on TI-89 Video
Raw Transcript
Hello again from everystepcalculus.com and everystepphysics.com. Another Limit to Infinity problem. Let’s do it. Index 8 to get to my menu. Choose Limits. Choose Compute Limit. Alpha before you enter anything in here. Alpha, parentheses, three times X plus two divided by x.
Add the infinity sign. Alpha yellow button here in the catalog which is the register for infinity. Show you what you’ve entered. I say it’s okay. We take all the terms and divided by the lowest order. X in the denominator, 3 0, the numerator limit is three. Denominator divided by Xis one. Answers is limit is three.
Calculus Q & A-Limits
Question:
Evaluate the limits:
lim(theta symbol–>0-) csc theta symbol
lim(theta–>pi/2) 1/2 tan theta
Calculate this limit at infinity (+/- infinity) for the following function:
f(x) = e^x+e^-x/2
Answer:
When you say f(x), calculus and programmable calculators are saying “y”. When they ask you for a limit, it is a trick question meaning find y. They then give you an x value and say as x approaches 5 (for instance) what is the limit? You then would plug 5 in for x ( which my programs do) and the answer is the limit is the value of “y”. Most of the time they trick you again do to the function they give you which is always a fraction such as (3x+7)/(x^2-4), (always a division sign). Then they ask you as x–>2 (as x approaches 2) what’s the limit. You’ll notice that when you plug in 2 for x you get a zero in the denominator which becomes an “undefined” answer.
If you then (like my program does) add some very small number to 2 such as .001 to get 2 + .001 = 2.001 or take a small number away from 2 to get 2-.001 = 1.999. Setting that equal to x then you actually get a number and that is the limit. I would actually graph the original function quickly on the calculator so I could see the picture of the function and then use the cursor to scroll back and forth at the x value they give you to do things like what happens when x approaches this from the right or left, to give me any clue of what their talking about.
Same thing with infinity. Graph the function first and then logically think about what they are asking. Also start to make things perfectly clear in your calculation examples so there is no question of what you’re asking.
For instance, you have:
f(x) = e^x+e^-x/2 which to a calculator, and me, means
f(x) = e^(x)+e^(-x)/2
and when you mean to say:
f(x) = e^(x)+e^(-x/2)
The problem says calculate the limit at infinity. I would graph the function on the calculator. You can see it’s just a parabola (valley looking, because no minus sign before it).
Here’s what they might be looking for:
f(x) = e^(x)+e^(-x/2)
Convert to:
= x*ln(e)+(-x/2)*ln(e)
factor out the ln(e)
= ln(e)[x+(-x/2)]
ln(e) = 1 So:
(1)[x+(-x/2)]
The limit as x-> ∞
= [ ∞ + (-∞/2)]
= ∞
Partial Fraction Decomposition Ex 11
Partial Fractions Example 3
Limits: Negative Infinity on TI-89 Video
Lim (7x^3+4x-17)/(4x^3-x^2)
Raw Transcript
Hello Everyone, Tom from everystepcalculus.com and everystepphysics.com. A problem regarding limits at Infinity. Index 8 to get to my menu. Scroll down to Limits. You can actually use second to go quicker on the menu, in other words, I want to go up a screen second, and up, second and up,or second and down. And here’s Limits. Click on that. And my other videos that I did previously I didn’t put Limit to Infinity here. You would access it by going to Compute Limit but when you want to put minus Infinity in there. It’s a little bit tricky do that. Well, it’s a little tricky on the Titanium to do that so I decided to put it in a actual menu. So here’s Limit to Infinity. We’re going to put the function in. Press Alpha first. Alpha left parentheses. This is a test problem. 7 times X cubed plus 4 times X minus 17, right parentheses, divided by, left parentheses, 4 times X cubed minus x squared, right parentheses. And let’s go negative infinity. Number two. And there it is there. This is the limit of this with X going to minus infinity. And I always show you if that’s correct. I always give you a change it in my programs. I say it’s okay. Then we’re going to divide all terms by the highest order term in the denominator. Which will be x cubed. Here’s the terms numerator. All divided by x cubed. Here’s the results. Now we put Infinity in for every X. Or minus Infinity, sorry. And that turns out Numerator Limit of 7. These turn out to be 0. Denominator
4 divided by the x cubed. That equals 4. And x squared divided by x cubed equals one over x. We add the minus Infinity to every x.
It turns out to be 40 Denominator Limit. The Problem Limit therefore is seven divided by four or seven fourths. Pretty neat, huh? everystepcalculus.com Go to my site and buy my programs. If you want to pass calculus and/or do your homework. And remember you’ll have these programs for life if you do that.
Limits Infinity Solved TI-89 Video
Raw Transcript
Hello Everyone. Tom from everystepcalculus.com and everystepphysics.com. Limits to Infinity. Let’s get started. Index 8 to get to my menu. Scroll to limits and choose number 2, compute the limit.And the function, you have to press alpha before you enter anything in here. Alpha left parentheses five plus two times X. This appeared on a test. Divided by,left parentheses, 3 minus X right parentheses.
Then press alpha and we press this yellow button here which goes to the letters with the yellow letters about the keys. The yellow register, I call it. And this is infinity right here. Alpha. There’s Infinity. I always show you what you’ve entered. The limit of two times x plus five over 3 minus X, as X approaches infinity. That’s the way you state it. I will show you what you’ve entered, you can
change it if you want. I say it’s okay. You divide all terms by the highest order term in the denominator. And that becomes a minus X
And you take the limited as that approaches infinity and the numerator limit is minus 2.Denominator, you have three and minus x in the denominator and that’s divided by minus x.Three minus x and you take the limit, you put Infinity in here. Anytime you take a number and divide it by Infinity, you get a very large number. It gets smaller and gets closer to zero. So that equals zero. And the other one equals one. That’s the denominator limit. You divide that numerator by the denominator, of course and you get minus two for the problem. Pretty neat, huh? everystepcalculus.com Go to my site and buy my programs.Thank you
Partial Fractions of Integral-Video
Raw Transcript
Hello Everyone, Tom for everystepcalculus.com and everystepphysics.com. I’m going to do partial fraction problem. This person said that it’s a hard partial fraction problem. Let’s see about that. It is hard if you ask me, without my programs. But let’s do it. Index 8 to get to my menu. Scroll down to partial fractions. We’re going to enter the integral. You have to press alpha before you enter anything in these entry lines. We’re going to go alpha and then going to do left parentheses, X squared plus 3 times X plus 1 close off the parentheses divided by, open up the parentheses, X to the fourth power plus 5 times X squared plus 4, close off the parentheses.
I always show you what you’ve entered. Now maybe you’re better than I am but you have to factor this denominator in partial fractions every time. So this it makes it hard for a class in it for a test problem is way too hard for a test probably for homework because I don’t know who would be able to write off from memory to factor that. I say it’s okay. We factor the denominator; here it is right here. And we start doing our partials. The idea is that you are going to deliminate the denominator here by multiplying at times the same thing with the numerator this. This here is the the factored part ion is here and this is the original denominator. And so then we have
the numerator is equal to Ax plus B times x square plus 4 plus Cx plus D times x squared plus 1.You multiply that out using the foil method. Remember the foil method? First outside inside and last. First would be Ax times x squared,outside would be Ax times four, et cetera. I multiply it out here. The calculator uses small letters rather than capitals. No problem. They combine those terms and come up with this. A plus C is x squared, x squared, 4a plus c, et cetera. Now we have to figure out the coefficients for this numerator. And since there’s no X cubed, we have to put one in there, which is 0 times x cubed. And then we have one coefficient here, three coefficient here, and one. So here we have one, three, and one. So then therefore AC is equal to 0, BD equal to one. 4a plus c is equal to three and 4b plus d is equal to one. And we work out, we use subtraction. We’re subtracting like terms. So AC is subtracted. 4a plus c and that equals three. A equals one. Do the same for each one of them. D equals one, C equals a minus one. Notice here we’re replacing A with what we found which is one. Which I do that for you here. When they’re in quotation marks, I replaced it in there. Partial fractions are this. Right here. And I also do the integrals for you. Log of x squared plus one divided by two, et cetera, et cetera. Pretty neat, huh?
everystepcalculus.com. Go to my site and buy my programs.
Thank you
Partial Fractions
Let’s talk about partial fractions and what I’ve found out after programming them. Partial fractions would never occur in real life. Remember the integral is the area under a smooth curve, nothing more from my knowledge, when you graph any original function; it has at least 2 asymptotes. Well that eliminates the smooth curve. Remember an asymptote is a vertical line or sometimes horizontal line where the original function never touches to infinity, so if you choose a range that goes (crosses) over that asymptote there is no computed answer, so no area under a smooth curve. So partial fractions are a Sudoku of math problem, like so many of calculus problems. So let’s get into what I’ve found in programming this area of calculus. The denominator has to be factored to produce partial fractions. Sometimes there are two factors, then three, then two factors with an exponent in between the parenthesis, then even one factor with an exponent outside the parenthesis. The factored denominator is the key as to how to approach the problem.
In my programs, I have a whole separate program to decide if there are two factors or three or any other choices. Most of us are not good at factoring. Maybe simple functions we can, like the difference of squares (x^2-9) = (x-3)(x+3). When it gets a little deeper that this most of us are lost. For instance, what’s the factors of (x^3-x) or (x^4+7x^3+6x^2)? The calculator knows and because of that, so does my programs. But would those appear in your test? In my opinion if they did then most of the class would fail that. No question about it. Let alone complete the partial fractions. So enough about factoring. Remember in algebra when they said you could do anything to one side of an equation as long as you do it to the other side also. They do this in partial fractions. On the left side of the equation in partial fractions they multiply the original function by the denominator, which effectively gets rid of the denominator. Well they do the same thing on the other side (except factored) to get rid of the added or subtracted fractions, and change the A / (x-1) to a product through common denominator. That said, a few trick to remember in partial fractions. If in the factored denominator you have (x^2+6) for the first factor so you want to set that up for partial fractions you would go Ax+B / (x^2+6), if the second factor was (x^2-3) you’d set this up as (Cx+D) / (x^2-3), Do you understand this? Very important and no exceptions!! If the factor turned out to be (x-5)^5(exponent outside of the parenthesis), this is called re-occurring powers, or exponents, and the factors are A/(x-5)^1 + B/(x-5)^2+C/(x-5)^3+E/(x-5)^4+F/(x-5)^5. Now if the power in the denominator is less or equal to the power in the numerator, you have to use short division to find the remainder to the partials.
My programs do all of this for you, as I required my programming to do for me also when I was in class. Hey good luck in your class, I’m always available to help you if you just ask.
Partial Fractions Example 10-Video
Partial Fractions | Example 9-Video
Raw Transcript
Hello Everyone, Tom from everystepcalculus.com and everystepphysics.com. Partial Fraction Decomposition. This time with three Partial Fractions.Let’s get started.Put index 8 in here with the open and closed parentheses, you get to my menu.You can scroll down to start up in the a section but you can scroll down alphabetical to Partial Fractions.Enter the integral by pressing Alpha first. Alpha, parentheses X squared plus 12 times X plus 12,close off the parentheses, divided by, open parentheses X cubed minus 4 times X. I always show you what you’ve entered. How many of you could factor this here? It’s pretty tough, I would think. Do it easy on my program. It comes up with 3 factors. So you write this in your paper. X squared plus 12x plus 12 is the numerator. Divided by the denominator times the denominator. And this is the actual denominator factored so we do the same thing here. And we’re going to use this to multiply times all of this to get rid of all the divisions. Common denominator is the actual words. So we added with x squared plus 12x plus 12 equals A and x times x minus 2 times x plus 2 Bx times x plus 2 Cx times x minus 2 And x equals 2. These quotation marks here is what the calculator does but you would put parentheses wherever you see these. parentheses 2 squared. You’re substituting 2 for every x in these functions. Use parentheses 2 here plus 12, etc. Do that throughout the whole system here. And that turns out to be B equals 5. At x equals zero. Use 0 for every x in all the computations here. And that turns out to be A equals minus 3 and x equals minus 2. Substitute for x in all the equation. C equals minus 1. Partial Fractions are minus 3 over x plus 5 over x minus two plus minus one over x plus two. And we integrate those. One thing about integrating a denominator with no, no exponent. You notice you can’t. This is an exponent of one, theoretically. But if you try to switch it to the numerator, five times X minus 2, it becomes a minus 1. Well in integration, you’re going to add one and then divide by the result. Well if you add one to a minus one, you get zero. In the numerator, anything to the zero is one. So that’s the reason you have to use logs. Whenever you see a no exponent to the denominator, it’s a log integration. So minus three logged minus the absolute X 5 log absolute x minus two minus one log x plus 2 plus C. Pretty neat, huh? everystepcalculus.com
Go to my site and buy my programs.
Thank you for watching.
Concavity-Video
y = x^3-6*x^2-3*x+1
Raw Transcript
Hello Everyone, Tom from everystepcalculus.com and everystepphysics.com. We’re going to do a problem on Concavity. And I’m gonna show you how my program work on that. Index 8 to get to my menu. We scroll down in the c section to Concavity. There is is there, letter D. I always have you start a graph on your paper. You put the x axis and y-axis with these. Keep up with graphing by hand, these functions. And I have a function here of the internet, for example. You have to press Alpha before you enter anything in these entry lines. Press Alpha. And the problem is X cubed minus six times x squared minus three times x plus one. I always show you what you’ve entered, you can check it and see if it’s correct. If it is, we press okay. I know we want Concavity, number 4. I’m going to press the number here, number four. Here’s Concavity. And I give the examples, here’s Concavity down and heres concave up. This is a valley, this is a mountain type situation. See, you take the original function. Take the first derivative here which is this. On your paper and the second derivative which is this. Six x minus twelve. We set 6 x minus twelve to zero to find what the x is at that which gives us critical number, there. And with respect to the second derivative and x equals two. So here we draw a number line here with the with two in the middle. Now we’re going to choose some number above two and some number below two. And and plug it into the second derivative here. I do that for you. I say X equals three. So at the second derivative x equals three, the answer is 6. And you’ll notice the 6 is positive. That means that its going to be concave up. And it’ll look like this on the graph. If we do a number less than 2 which is one. And plug it into the second derivative and we get minus 6. So, this is negative. So this is gonna be down. And I show you this and the next screen. So this one is down when you when you graph this function, it’s going to be down like this is going to be up like. That’s called Concavity. And that’s how you do it . Now on your paper, you must write negative infinity to two and two to positive infinity. And then you’ll be correct in your test or homework. Pretty neat, huh?
everystepcalculus.com
Thank you
Chain Rule Example Solved-Video
Raw Transcript
Hello everyone. This is Tom from everystepcalculus.com.
I’m going to demonstrate the Chain Rule right now.
Let’s do it. Index eight to get to my menu for calculus.
And we’re going to
scroll down to the C section. They’re all alphabetical.
Here’s chain rule here.
And I gave you a choice of what kind
problems you’re gonna do. Sine, Cosine, Tangent, etc.
Or given Y, etc.
Here’s the, in parentheses, with an exponent.
And here’s the square root of something. I’m going to do number 2 here.
Press alpha before you
put anything in these entry lines here.
Alpha first and the left parentheses which it gives you up here for example.
And we’re going to 5 times x squared
plus 7 to the fifth power
I always show you what you entered. You can change it if you want.
We’re going to differentiate that.
I say it’s okay. Here’s the formula here for chain rule.
and I go through the steps for you. It’s kinda like U-Subsitution
in a way for differentiation. U-Subsitution is really for integrals.
So after we put the terms into the into the formula. Mark everything down in your paper you see it.
Here’s the derivative of g of x
which is 10 times X
and keep writing and here’s the answer.
Solving Definite and Indefinite Integrals – Q&A
Dear Tom,
I would like to know if your program also solves integrals consisting only of symbols like x, a or b in both definite and indefinite matters?
Example:
or
Would solving does definite integrals be possible using your programs showing the solution process step by step?
Kind regards,
Ben
Answer
The first integral, sounds like your professor didn’t teach you, so you will want my programs to teach you or me to teach you. A common endeavor in my experience. When you see the da that’s “with respect to” in other words, you are integrating that integral with respect to “a”. In my programs you’d change all the “a’s” to x”s.
Now could you solve the integral if it looked like:
x x
∫ (1/x)dx or ∫ ( 1/[(c+b)-(k+x)] ) dx
o o
Now these would never appear on a test unless your professor demonstrated them on the black board, so they are either homework or experimenting off the internet. Both integrals are definite integrals because they show a range to integrate over. When you solve the integrals you have the area under that functions curve. That’s all integrals do is solve area under curves. You can graph 1/x on your calculator and you can see what the first integrals function looks like. Now, anytime you see a function without an exponent in the denominator, it is a natural log answer. The reason is is that you can’t integrate term with a division sign.
In this case the x has an exponent of 1 ( x^1), you always have to move the denominator to the numerator to eliminate the division sign before you can integrate. If you move the x^1 to the numerator to eliminate the division sign it becomes x^(-1) so when you do the integration process which is always add 1 to the exponent and divide the term by that answer, However when you add 1 to a -1 you get zero, and any term to the zero exponent = 1, so that’s why the natural log comes in. In the first integral above = ln(x), then you can enter the ranges for x that you are given, do the top range first and subtract the bottom range to get the answer (area under the curve of the function). The second integral above you’d break it into two integrals ∫[1/(c+b)] dx which equals x/(c+b) and then the second term 1/[(k+x)]dx which equals ln(k+x).
U-Substitution using Log Rule on TI-89
U-Substitution using Log Rule on TI-89
Raw Transcript
Hello everyone I’m Tom from every step calculus dot com.
Im gonna do a U-substitution problem
that requires the log rule. And uh,
let’s get into it. index8() is my,
you have to put that in the end line here to get to my menu.
I’m already at U-substitution.
You scroll down with the cursor to get there.
And wait for it to load here for a second.
And we’re gonna enter our function. You have to press alpha before you enter anything in my
entry lines in my programs.
Alpha 8 times
X divided by
parentheses
X squared plus
:00
I will show you what you’ve entered. The reason you need to do a log rule is
notice in the denominator your have x squared plus one.
This is an exponent of 1. If you we’re to
transfer this up to the numerator the exponent of
one would become a -1. And in integration you always add,
the first thing you do is always add to the numerator.
And one plus minus 1 is zero so the answer will always come out to be
-1. That’s the reason anytime you see this without, without an exponent other than
one. The one is of course hidden. Um,
you know it’s a log problem.
So we’re going to press there.
Say its okay. Here’s the original
function. And all these you have to rewrite them. Notice there is a constant error
of 8. You have to bring that outside the integral, that’s very important. And we’re going to also
take the x
here and put it over by the dx.
And we’re going to put one in the numerator,
for X squared plus 1 in the denominator.
So U is x squared plus 1 the derivative of that is 2x.
Another trick to all these, took me a long time to figure out.
Was d you need to have the divisor of 2
You have to move that over as a divisor here, and leaving
this here. You’ll notice that X DX is the same as x dx.
That means it’s a U-substitution problem if it wasn’t it’s not a
U-substitution problem. I ask you that though just
to make sure you know what you’re doing. I say yes.
So we have i8 outside the integral and one over you
du number 2. Now we have a constant here at one half.
So we have to bring that outside the integral, which we do over here.
8 and 1/2 that equals 4.
And then we have four times log of U. Anything over one of U is
log of u, plus c.
And the answer is, after we substitute the u back in.
4 times log of x squared
plus 1 plus c. Pretty neat huh?
Every step calculus dot com. Go to my site buy my programs, pass your calculus
class and
have these programs for the rest of your life your kids or grandkids or
your sister, whoever might take calculus in the future,
you have these forever, that’s what’s good about them. And also subscribe to me on my
you know, so that you can see the future
movies or the other blog.
Partial Fraction Decomposition on TI-89
Partial Fraction Decomposition on TI-89
Raw Transcript
This is a video from EveryStepCalculus.com demonstrating how my programs work on a TI-89 Titanium calculator and other calculators in the TI system for physics and calculus problems.
Okay, Partial Fraction Decomposition. Let’s get started, to get to my menu you have to press 2nd alpha to put the i_n_d_e_x in here you have to press alpha to put the 9 (). Press Enter and you’re into my menu. Scroll down to wherever you want all the way up and down.
But I’m at Partial Fractions right now which we want to do and we’re going to enter our function. You always have to Press alpha before you enter anything in a these entry lines in my programs. So Alpha we’re gonna do a quantity of 4 times X plus 9, divided by the quantity X minus 1, times the quantity of x plus 1, times the quantity of X plus 4.
I always show you what you’ve entered so you can change in case you made a mistake and we’re into the problem. You factored the denominator. It’s already factored here but in case it wasn’t, I would. And then you multiply, you multiply by the factor of the denominator. You get these. They’re all exactly right. Mark it down on your paper exactly like you see it. And if x = 1, the idea is that you cancel certain factors, certain variables. And here’s when you put -1 the function here is where they all set.
Mark that on your paper, turns out to be 5=A*(0)+B*(-6)+C*(0) 5=B*(-6) 5/-6. Mark all that on your paper. X equals one which will negate the second term. You mark these all these… put all these in replace or um x in your function there. A couple of them become zeros so that you can solve so a equals 13 tenths. And then you have enough variables, enough answers to solve the other c. Here’s the answer.
Pretty neat, huh? EveryStepCalculus.com. Go to my site, buy my programs and pass calculus.
More Partial Fraction test questions solved on your TI-89 calculator
Partial Fraction Test Question #1
U Substitution e^(x) Solved on TI-89
U Substitution e^(x) | Every Step Calculus
Raw Transcript
Hello Tom from every step
Calculus dot com. Here’s a problem on u-substitution.
L,et’s get started index8() to go to my menu.
u-substitution is the subject.
Press alpha before you enterr anything in these entry lines here.
Diamond key to get to the
e(x) problems this is 7 times
X. Divided by.
Anytime you using the denominator you always have to have parentheses
around the whole function.
And then we’re going to go
to e(x) again. 7
times X plus
8.
I always show you what you’ve entered, you can change it if you want, I say it’s okay.
You have to rewrite this. Notice the idea is that the seven ex/dx is the same as
here. Otherwise it’s not a u-substitution problem.
And then I check whether that’s that’s true.
So they’re equal so we say yes.
And here’s the answer 7 times log e to the x
plus c. Pretty neat huh? Every Step Calculus dot com, go to my site buy my programs
pass calculus, also subscribe to
my channel so you can see more movies
or blogs.
3 Calculus Questions Solved by TI-89
I’m really trying to figure out how to use your program with a couple different problem examples:
“Find the following derivatives of f(x). Do not leave any negative exponents”
1. F(x)= 4^x-ln(2x^2-x)
What program can I use to solve this?
another is f(x)=2csc(x)+5arccos(x)
This is fundamental to calculus and you must start to learn it or know it.
Notice the minus sign between the terms (if a plus no difference)
This means that the derivatives of the terms are separate or individual
You differentiate each by itself and one at a time
There could be 15 more terms each separated by a minus or plus and you’d do each one separate.
So: f(x) = 4^x f’(x) = x*4^(x-1) Notice when differentiating you take the exponent
Multiply it times the front of the term and take one away from the exponent.
You must do this process in your sleep and fast or you’ll have trouble passing calculus
Another Example
f(x) = 15x^7
f ‘(x) = derivative
= 7*15*x^(7-1)
= 105x^6
Integration is the exact opposite
You add one to the exponent first
Then divide by that new exponent
∫[105*x^6] dx
= 6+1 = 7
So: 105*x^7 / 7
= 15x^6
Problem 1) the trig problem
Reload the attached program
First Delete the same named program from you calculator
Then reload the attached changing to “Archives” first
For the problem go to:
“trig” in my menu
Scroll to csc(x)
Find the derivative and add the 2 in the front of the answer
For arcos(x)
Scroll to cos-1 in the menu of trig
And find that derivative adding the 5 in front
“Answer the following questions about the given function: f(x)=x^3-6x^2+9
A) Find f'(x)
B) Find f”(x)
These again are separated by minus and plus signs so each is differentiated separately
f(x) = x^3 – 6*x^2 + 9
f ’(x) = 3x^2 – 12x + 0
f ‘’ (x) = 6x – 12
Notice I took the exponent and multiplied it by the front of the term
And then took one away from the exponent in each case.
Do these in your sleep!!!!!
Use log differentiation to find the derivative of the following function with respect to variable x.
(x^3+2)^3sqrt(2x^2+3)
Notice the “times” sign so you’d be thinking “product rule” or logs to differentiate
First it must be expanded:
I have expanded log problems to another base
But haven’t programmed to expand natural logs, but am in the process.
Should be done soon
(x^3+2)^3 * √(2*x^2+3
ln((x^3+2)^3) + ln(√(2*x^2+3) (always you change a square root to ^(1/2)
= ln( (x^3+2)^3 ) + ln( (2*x^2+3)^(1/2) )
= 3*ln( x^3+2 ) + (1/2)*ln( 2*x^2+3 )
Now you can choose in my menu
“log Problems”
Choose “ln(x)”
Choose “differentiate”
And add the above ln tems one at a time
Put the problem back together when finished with the individual solutions.