# Higher derivative product rule second degree differentiation

• 6 Oct 2017, 6:44 PM
Thatstudentm9
Higher derivative product rule second degree differentiation
Hey guys I was doing this question in the following methods:

question: find d^2v/dt^2 if v=(t+3)(2t-1)2

1) u=t+3 u'=1 and v=(2t-1)^2 v'=4(2t-1)

2) sub in the values obtained into u'v+v'u ................................ =1(2t-1)^2+4(2t-1)(t+3)

3) (from this point on i was somewhat confused whether to expand and then factorise or just jump st8 into factorization)

this btw the first degree differentiation moviand reng of to the second. I know this is a long question requires heaps of working out but can someone please help me with this question as i saw it in a past paper and csa trial paper. Your help would be highly appreciated
• 6 Oct 2017, 7:17 PM
InteGrand
Re: Higher derivative product rule second degree differentiation
Quote:

Originally Posted by Thatstudentm9
Hey guys I was doing this question in the following methods:

question: find d^2v/dt^2 if v=(t+3)(2t-1)2

1) u=t+3 u'=1 and v=(2t-1)^2 v'=4(2t-1)

2) sub in the values obtained into u'v+v'u ................................ =1(2t-1)^2+4(2t-1)(t+3)

3) (from this point on i was somewhat confused whether to expand and then factorise or just jump st8 into factorization)

this btw the first degree differentiation moviand reng of to the second. I know this is a long question requires heaps of working out but can someone please help me with this question as i saw it in a past paper and csa trial paper. Your help would be highly appreciated

$\noindent Maybe quicker to just use (fg)'' = f''g + 2f'g' + g''f. One of the terms will automatically vanish because the second derivative of the linear factor is 0.$
• 6 Oct 2017, 8:00 PM
Thatstudentm9
Re: Higher derivative product rule second degree differentiation
Quote:

Originally Posted by InteGrand
$\noindent Maybe quicker to just use (fg)'' = f''g + 2f'g' + g''. One of the terms will automatically vanish because the second derivative of the linear factor is 0.$

I'm not familiar with this method as i've never even heard of it could please elaborate on this method whats g btw????
• 6 Oct 2017, 8:26 PM
InteGrand
Re: Higher derivative product rule second degree differentiation
Quote:

Originally Posted by Thatstudentm9
I'm not familiar with this method as i've never even heard of it could please elaborate on this method whats g btw????

It's what you get from applying the product rule twice.
• 7 Oct 2017, 1:44 PM
jjHasm
Re: Higher derivative product rule second degree differentiation
Quote:

Originally Posted by Thatstudentm9
Hey guys I was doing this question in the following methods:

question: find d^2v/dt^2 if v=(t+3)(2t-1)2

1) u=t+3 u'=1 and v=(2t-1)^2 v'=4(2t-1)

2) sub in the values obtained into u'v+v'u ................................ =1(2t-1)^2+4(2t-1)(t+3)

3) (from this point on i was somewhat confused whether to expand and then factorise or just jump st8 into factorization)

this btw the first degree differentiation moviand reng of to the second. I know this is a long question requires heaps of working out but can someone please help me with this question as i saw it in a past paper and csa trial paper. Your help would be highly appreciated

im assuming theres a little typo and that v is meant to = (t+3)(2t-1)^2. If the way InteGrand mentioned seems too confusing, just expand everything and differentiate v 2 times, so u wont have to worry about product rule. Just simple differentation, no brackets. I believe it should be 24t+16 ? All InteGrand is doing I believe is using product rule 2 times?

Either this or im retarded and don't know what the question is asking for :P
• 8 Oct 2017, 5:41 PM
braintic
Re: Higher derivative product rule second degree differentiation
Quote:

Originally Posted by InteGrand
$\noindent Maybe quicker to just use (fg)'' = f''g + 2f'g' + g''f. One of the terms will automatically vanish because the second derivative of the linear factor is 0.$

I'm quite sure HSC markers won't allow this. If you are going to use formulas from beyond the course you have to derive/justify them first.