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# Maths C3 Help watch

1. 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!
2. (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.
3. (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 )
4. (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?
5. (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
6. (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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