Math Is Fun Forum

ninjaman

Member

Registered: 2013-10-15

Posts: 61

derivatives

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)

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: derivatives

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.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

ninjaman

Member

Registered: 2013-10-15

Posts: 61

Re: derivatives

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?

ninjaman

Member

Registered: 2013-10-15

Posts: 61

Re: derivatives

nope!

nevermind!

I got it!

I HAVE GOT THIS MAN!!!!

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: derivatives

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.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.