# Thread: Higher derivative product rule second degree differentiation

1. ## 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

2. ## Re: Higher derivative product rule second degree differentiation

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

3. ## Re: Higher derivative product rule second degree differentiation

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????

4. ## Re: Higher derivative product rule second degree differentiation

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.

5. ## Re: Higher derivative product rule second degree differentiation

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

6. ## Re: Higher derivative product rule second degree differentiation

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.

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