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**ninjaman****Member**- Registered: 2013-10-15
- Posts: 41

I have to do second and third order derivatives

(2x-3)^4

for the first one I got

8(2x-3)^3

for the second I got

48(2x-3)

the way I was shown is this,

2x-3=U du/dx = 2

y= u^4 dy/du = 4u^3

dy/du * du/dx = 4u^3 * 2 = 8u^3

since u = 2x-3

8(2x-3)^3

then I got lost

here I got stuck and not sure where I went wrong, any help

thanks

simon

*Last edited by ninjaman (2014-01-22 09:40:20)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,429

Hi;

for the second I got

48(2x-3)

That is incorrect.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**ninjaman****Member**- Registered: 2013-10-15
- Posts: 41

how do you do the third with that, also I checked an online derivative calculator and it says that is wrong?

it gave the same answer with the power of 2, 48(2x-3)^2

I looked at the steps on this calculator and didn't understand them.

Im not sure how to go onto the third derivative.

would I use (2x-3) as U or 48(2x-3) as U?

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**ninjaman****Member**- Registered: 2013-10-15
- Posts: 41

nope!

nevermind!

I got it!

I HAVE GOT THIS MAN!!!!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,429

Very good. The easiest questions to answer are the ones where the questioner figures it out.

Im not sure how to go onto the third derivative.

would I use (2x-3) as U or 48(2x-3) as U?

Use the (2x-3) for u. Constants really do not figure in the process. Just do not forget to hold on to them for the final answer.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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