Can Calculus Problems be Solved by Factoring?

[carl binney]

Homework Statement

[/B]

hi could some body please help me factorise this please ? any chance of a few stages would be much appreciated

Homework Equations

The Attempt at a Solution

my attempt , but my solutions say otherwise ?

[/B]

 

[Mark44]

Homework Statement

View attachment 81835 [/B]

hi could some body please help me factorise this please ? any chance of a few stages would be much appreciated

Homework Equations

The Attempt at a Solution

my attempt , but my solutions say otherwise ?
View attachment 81836 [/B]

The 2nd line in your attempt is wrong. If you had posted the work itself rather than an image of your work, I could show you exactly where you went wrong. That's why we insist that you post your work directly here, not just post an image.

 

[SammyS]

Homework Statement

[/B]

There really is no problem statement here, just a mathematical expression which looks suspiciously like it's the the result of a botched attempt at applying the product rule, and perhaps the chain rule.

 

[BvU]

Dear Carl,

I get the impression you didn't read the guidelines as I asked you to do in your first thread. The two replies above should indicate to you that a different approach is in order: after all, if you don't want to be bothered with typing out something simple like

6(x+3)⁶ (2x-5)² + 6(x+3)⁵ (2x-5)

then why should a PhD or MSc who makes his/her precious time available to help others bother to even read your postings ? (which, to be sure, are at least upright and clearly legible -- I've seen a lot worse).

Back to business: as Mark wrote, your second line is a re-write of only the first term (with a 6 too many)

 

[carl binney]

hi sorry about the mistakes, the question is

differentiate , y = (2x-5)^3 (x+3)^6

I can get up to where you have to factor the equation, but am unsure what to do from there , any help would be much appreciated ??

 

[SammyS]

hi sorry about the mistakes, the question is

differentiate , y = (2x-5)^3 (x+3)^6

I can get up to where you have to factor the equation, but am unsure what to do from there , any help would be much appreciated ??

Do things a step at a time. Differentiation (taking derivatives) is a very methodical process.

First use the product rule.

What form of the product rule do you use?

 

[carl binney]

= (x+3)^6 * (3(2x-5)^2*2) + (2x-5)^3 * (6(x+3)^5)
= (x+3)^6 *(6(2x-5)^2) + (2x-5)^3 * (6(x+3)^5)
= 6(x+3)^6 (2x-5)^2+ 6(2x-5) (x+3)^5
factoring this part is where I get stuck ?

 

[Mark44]

differentiate , y = (2x-5)^3 (x+3)^6

= (x+3)^6 * (3(2x-5)^2*2) + (2x-5)^3 * (6(x+3)^5)
= (x+3)^6 *(6(2x-5)^2) + (2x-5)^3 * (6(x+3)^5)
= 6(x+3)^6 (2x-5)^2+ 6(2x-5) (x+3)^5

You have a mistake in the line above. The exponent on 2x - 5 in the 2nd term should be 3, not 1.

factoring this part is where I get stuck ?

From your last correct step you have y' = 6(x + 3)⁶(2x - 5)² + 6(2x - 5)³(x + 3)⁵.

You have two terms here, the large expressions that are connected with '+'. Both terms have a factor of 6, so that will be part of the greatest common factor (GCF). Both terms have at least one factor of (x + 3). The GCF will have (x + 3) to the smaller of the two powers that are present in the two terms.

Also both terms have at least one factor of (2x - 5). The GCF will have (2x - 5) to the smaller of the two powers that are present in the two terms.

After you figure out what the GCF is, factor it out of both of the large terms. The other factor will consist of two simpler terms (connected with '+'. Your earlier work was incorrect, in part, because it was missing the '+' between the remaining terms after you factored out the GCF.

As an example, suppose you have $6x^3y^2 + 12x^2y^4$. Here we have two terms. The GCF has 6 in it, because 6 divides the coefficients of each of the two terms. One term has x³ and the other term has x². The smaller of these two exponents is 2, so the GCF will include x². Finally, the first term has y² and the second term has y⁴. The smaller exponent is 2, so the GCF must include y².

From this, we see that the GCF is 6x²y². We proceed as follows:
$6x^3y^2 + 12x^2y^4 = 6x^2y^2(x + 2y^2)$
As a quick check, expand what is shown on the right, and you should get the expression on the left side.

 

[epenguin]

hi sorry about the mistakes, the question is

differentiate , y = (2x-5)^3 (x+3)^6

I can get up to where you have to factor the equation, but am unsure what to do from there , any help would be much appreciated ??

= (x+3)^6 * (3(2x-5)^2*2) + (2x-5)^3 * (6(x+3)^5)
= (x+3)^6 *(6(2x-5)^2) + (2x-5)^3 * (6(x+3)^5)
= 6(x+3)^6 (2x-5)^2+ 6(2x-5) (x+3)^5
factoring this part is where I get stuck ?

WHERE, where, where?
The factorisation is very obvious to me, RHS of results has the same bracketed terms just to different powers appearing in each of the terms that are added, so each of those terms.to the lower power is a factor of the whole.
However it is true that not only logic but familiarity and practice count in math so this may not be obvious to you.

So make it easier for yourself.

You must have heard in mathematical discourses the term 'WHERE' used. A long complicated mathematical expression is written shorter, saying 'WHERE' some letter is equal to something more complicated, such as w = some complicated expression or even something more implicit, such as 'WHERE w is the solution of ... some other equations.

This is because, contrary to what beginners tend to believe there is no virtue in complication, on the contrary. So they, the mathematicians and math users, use all this where, where, where to enable focussing the essentials of an argument.

Thus you might have seen easier the essentials - and the factorisations - if you had posed the problem as

Differentiate y = p³q⁶ where...
or better y = (pq²)³. where...

I leave you to fill in and use the ...

 

[epenguin]

This is calculus, not precalculus.

 

[Mark44]

This is calculus, not precalculus.

What the OP is stuck on is not the actual differentiation, but the factoring, so I did not move this thread.

 

[epenguin]

OK and now I think of it some of the more difficult bits of calculus are really algebraic problems.