well in this video, we may have the longesttitle of the whole page of videos here, applying laws of exponents and distributive rule towrite expressions in forms convenient for calculus. what does that all mean? first ofall, the first examples we look at are going to be in the form of a, where a is just somenumber, x is a variable, and n is another constant, or the exponent. ax^n is a convenientform for calculus. it's one of the simpler types of functions that you encounter. andso we want to be able to practice in writing things that can be written that way. so firstof all, what do i mean? here's some examples, two x to the seventh. in that case, a wouldbe two, and n would be seven. ok, three halves x to the negative five, in this case, a wouldbe three halves, and n would be negative five.
so that's the idea. we want to get something,we want to start with whatever expression we're given and see if we can write it inthis form, ax^n. so the first example is five divided by two x cubed. so that's the expressionthat we have, and we want to write it in our convenient form of a number times x to somepower. well, when you look at this, you need to or one way to think about this is to recognizethat this is a fraction, but we could also think of this as being made up of other simplerfractions. for example, this five in the numerator, we could think of that as being five overtwo, and then the other fraction that we would need to multiply by that to get five overtwo x cubed would be one over x cubed. now that's not the only way that you could splitit up, but i'm splitting it up that way because
it's going to get me to where i'm trying togo, that is ax^n. so right away, you can see we have a number a here, five halves, andif we could just write this one over x cubed as just x to some power, we'd be done. andthat's exactly what you learned in the last module, or perhaps you knew this already.one over x cubed is x to the negative three power. and so now, we have a as five halvesand the n as negative three. and we have ax^n. ok, so, and when you are in calculus, you'llfind out that if you can write something in this form, ax^n, you can work with it a lotmore easily than in the original form. now second example, this time let's take sevenover the cubed root of x squared. and i want to write it ax^n. now again, i'm still thinkingabout it, you don't have to write all of this,
but i'm still thinking about it as i wantto separate my numbers and my powers of x. so i have seven over one multiplied by oneover the cubed root of x squared. so now the seven over one is just seven. and so i canjust say seven and one over the cubed root of x squared. now there's two steps in thethought process here. the cubed root of x squared, we saw before, is x to the two thirds.so i'll put in the two steps. so the first step is to recognize that the cubed root ofx squared is x to the two thirds. so i have one over x to the two thirds. and then oneover a power of x is the same as the power with the exponent changing sign. so this wouldbe x to the negative two thirds. seven x to the negative two thirds, again we've got thea, the x, and the n. n is negative two thirds
in this case. now each one of these examples,if you want to get the most out of this, now that you've seen an example or two, you wantto pause, there's a pause button on the video, you should pause and see if you can work itout and then see if you agree with what we end up with in the video. so, next one, oneover the square root of seven x. seven x is under the square root sign. there's a coupleof ways to go about doing this. again, you should pause before we do this and see ifyou can come up with something. now one thing i guess, one way to look at this is that we'vegot a one over something. and so if that something was a power, then we could just use our lawsof exponents. so for example, we know the square root is, in this case, the one halfpower. so this would be seven x raised to
the one half power, so one over that. nowone over that would just be seven x raised to the negative one half power. now, you say,"wow, we're done. a is seven and n is negative one half," but not quite yet because the sevenis not really just a seven. it's inside these parentheses, and that parentheses has a negativeone half exponent on it. but we have a law of exponents that says that if you have differentbases and you raise them to a power, you can raise each base to that same power. and so,that would be seven to the negative one half multiplied by x to the negative one half.now seven to the negative one half power is a number. it's one over the square root ofseven. so you could write it like that, if you wanted to, one over the square root ofseven, and that's a number, and then we have
our x and we have an exponent on the x, inthis case negative one half. and so now we have a, one over square root of seven, timesx, and x is raised to the negative one half power, so n is negative one half. so that'sone way to leave the answer. you should not have stopped though with the seven x to thenegative one half because it's specifically the form is a number times just an x to apower, in this case negative one half. so that's the third example. now, this is theform a x to the n and like i said, that's the convenient form in calculus. it's nottoo often that you'll get something that can just be written in this form. sometimes youhave to solve for a sum of terms that are in this form. so i want to write in the nextexamples, write the expression as a sum of
terms in the form a x to the n. so we're going to have several terms addedtogether, a sum. now sum and difference, it's ok to have subtraction so that's just a sumwith a negative coefficient. so the form though is we have more freedom because now we canwrite it as a sum of terms. now, as a first example, and again, don't forget to pause,first example, suppose we have the square root of x times the quantity three plus xplus x squared. and we want to write that as a sum of terms. well you might say, i seea sum of terms, three plus x plus x squared. it is a sum of terms but it's multiplied bysomething. and we just want to have a sum of terms, each term being in the a x to then form. now, so we want to have some things added together here. and each one of thosethings has to be in our convenient form. so
the thing that comes to the rescue here isthe distributive rule, and that's where that came into the title. distribute the squareroot of x through the parentheses. and when we do, we'll have three square root of x plussquare root of x times x plus the square root of x times x squared. so that's the distributiverule. and that gets us to a sum of terms, which is what we wanted. we have a sum ofthree terms. now we just have to make sure that each term is in the form a x to the n.but that's not very hard in this case. the first one is in that form, if we just rememberthat the square root of x is x to the one half. and then remember the laws of exponents.when you multiply powers of x, you add exponents. so this would be x to the one half multipliedby x to the one. and so that would be x to
the one half plus one which is x to the onehalf plus two halves is three halves. so that would be x to the three halves. and then wehave the first term again. and then we have to go to the third term. and the third termis the same type of thing except we have x to the one half multiplied by x to the two,two is four halves, so that would give us x to the five halves. and so now we have thesum of terms and each term is in the form a x to the n. in the first case, a is threeand n is one half. in the second case, a is one and n is three halves. and the third case,a is one and n is five halves. so we've got terms that are in the form that we're callingconvenient for calculus. now second example, again pause and try it yourself before youlisten to the rest of the, my babbling on.
so, let's see, the next example is x squared plus three over three x to the fifth.now i want to write it as the sum of terms. and the terms have to be in our special forma x to the n. there's a couple of ways to look at this. the way i like to think aboutthis is i have to remember a little algebra here. it's very simple algebra that is youhave a plus b divided by c, the whole quantity a plus b divided by c, that you can thinkof that, if you want to, as the sum of two fractions, a over c plus b over c becausewith the common denominator of c, if you added them together, you just have a plus b quantityover c. so that's a very simple algebra fact that you've known a long time. but that'sexactly what we can use here. so in this case, the a would be the x squared and the b wouldbe the three and the c would be three x to
the fifth. so we'd have x squared over threex to the fifth plus three over three x to the fifth. having done that, now we just tryto simplify what we have. so the first term, the x squared will cancel with the x to thefifth. that will leave me three factors of x in the denominator. so the first one simplifiesto one over three x to the third power. and then the next one, the three's cancel, andwe have just one over x to the fifth. so we're getting there. now it looks like we're almostthere. we just need to write this as a number times power of x. in this case, the numberwould be one third, coefficient would be one third. the power of x, we've got one overx cubed. so one over x cubed would be x to the negative three power. so that's the firstterm. the second term is even easier. it's
just x to the negative five. and so, again,we have a sum, there's the sum, of two terms. and each term is in our convenient a x tothe n form. now the next example, example three, i suppose it is, in this section, it'sx divided by x squared plus three. so now we have to write this as a sum of terms. andeach of the terms is an a x to the n. so, i guess the first thing you try to do is thinkwell maybe i'll use my algebra again. c over a plus b. and that's c over a plus c overb. now is that true? i mean it kinda looks good. but if you were going to add these fractionson the right hand side, you would have a common denominator of a times b, not a plus b. andthen you would have bc plus ac in the numerator, and you would not get back what you startedwith here. now you can also convince yourself
it's not true by just taking almost any numbersfor a, b, and c, and it probably won't work out. stay away from ones and zeroes. but ifyou take two for c, three for a, and four for b, you'll probably find out that thisis not the same. so that's enough to prove it doesn't work as well. but this is not true.so put a big cross through that equals. that's not true. so that really keeps us from takingthe easy step here. we could have said x over x squared plus x over three, and we'd be homefree, except it's just not true. so well you say, well, maybe i should try something else.maybe if i think of this whole denominator as x squared plus three all to the one power.and then i could bring it up and say x times x squared plus three to the negative one power.that's looking promising. it's looking more
like what we want. but now we run into anotheralgebraic problem. and the question is so what can you do about this quantity to thenegative one power? can you say it's x to the negative two plus three to the negativeone, question mark? is that true? well times the x, i'm just working on the quantity inparentheses. well again, if you tried to add the this back together, if you got a commondenominator, you won't get back to where you started. and again, if you took a number forx, say x equals one, then you would have one plus three is four, four to the negative onepower is one fourth, but here you'd have one to the negative two which is just one, andthree to the negative one which is one third, one plus a third is four thirds. it's notone fourth. so this is not true either. so
don't do it. now, so, we've tried two thingsthat look somewhat promising and neither one of them worked. now perhaps then, well maybe,i started to say maybe we could take x over x squared and x over three, but that's thesame thing i said over here, with the c's and the a and the b. so i guess after youstruggle with this awhile, hopefully, you don't do any of these things that i've justsuggested, because they are not right. and so at that point, you just wonder well whatcan i do. and the answer is this can't be written in our convenient form. so the wisething to do here is to just punt. you can't do it in the form we want. sometimes you can'twrite the expression you have in the convenient form a x to the n, or even as the sum of ax to the n type terms. and so in that case,
you will have to learn some other things.and that's why calculus is a five hour course. so hope this helps and good luck in your calculus.
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