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    This is question 7b from AQA C3 Jan 2010, I can't insert a picture but this is the link to the paper http://filestore.aqa.org.uk/subjects...W-QP-JAN10.PDF

    I've done part a) fine, but don't know how to do part b). I'm probably missing something really obvious but the solutions I've looked at have suddenly got d2y/dx2 = 2tan4x x 4sec^2(4x) from somewhere, i just can't work out where?

    Thank you if you can help me!
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    (Original post by ksma1999)
    This is question 7b from AQA C3 Jan 2010, I can't insert a picture but this is the link to the paper http://filestore.aqa.org.uk/subjects...W-QP-JAN10.PDF

    I've done part a) fine, but don't know how to do part b). I'm probably missing something really obvious but the solutions I've looked at have suddenly got d2y/dx2 = 2tan4x x 4sec^2(4x) from somewhere, i just can't work out where?

    Thank you if you can help me!
    Correct, but you need it in terms of y, remember y=tan4x. So get all functions in terms of tan4x and substitute y for tan4x.
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    (Original post by ksma1999)
    This is question 7b from AQA C3 Jan 2010, I can't insert a picture but this is the link to the paper http://filestore.aqa.org.uk/subjects...W-QP-JAN10.PDF

    I've done part a) fine, but don't know how to do part b). I'm probably missing something really obvious but the solutions I've looked at have suddenly got d2y/dx2 = 2tan4x x 4sec^2(4x) from somewhere, i just can't work out where?

    Thank you if you can help me!
    The derivative is p(1+tan^2 4x), which is p(sec^2 4x). The second derivative is the derivative of p(sec^2 4x), which is p(2sec^2 4x x tan 4x )
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    (Original post by h3rmit)
    The derivative is p(1+tan^2 4x), which is p(sec^2 4x). The second derivative is the derivative of p(sec^2 4x), which is p(2sec^2 4x x tan 4x )

    But how do you know that that's the second derivative? It's not in the formula book?
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    (Original post by ksma1999)
    But how do you know that that's the second derivative? It's not in the formula book?
    It's pretty easy to differentiate.
    Write it as (cos A)^-2, use the chain rule/bracket rule (whatever you call it), get (-2)(-sin A)(cos A)^-3

    The negatives cancel, you get one tan and two sec^2's. Probably simpler than using the quotient rule again
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    (Original post by h3rmit)
    It's pretty easy to differentiate.
    Write it as (cos A)^-2, use the chain rule/bracket rule (whatever you call it), get (-2)(-sin A)(cos A)^-3

    The negatives cancel, you get one tan and two sec^2's. Probably simpler than using the quotient rule again

    Okay thank you for your help, I'll try that now!
 
 
 
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