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Thread: Higher derivative product rule second degree differentiation

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    Exclamation 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

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    Re: Higher derivative product rule second degree differentiation

    Quote Originally Posted by Thatstudentm9 View Post
    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
    Last edited by InteGrand; 6 Oct 2017 at 8:26 PM. Reason: Typo
    jjHasm likes this.

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    Re: Higher derivative product rule second degree differentiation

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

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    Re: Higher derivative product rule second degree differentiation

    Quote Originally Posted by Thatstudentm9 View Post
    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.
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    Re: Higher derivative product rule second degree differentiation

    Quote Originally Posted by Thatstudentm9 View Post
    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
    Last edited by jjHasm; 7 Oct 2017 at 1:44 PM. Reason: typo

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    Re: Higher derivative product rule second degree differentiation

    Quote Originally Posted by InteGrand View Post
    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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