AQA JAN 2010 PC3 Question 7(b) help

Could anyone help with this question (7b). I'll link it here (http://filestore.aqa.org.uk/subjects/AQA-MPC3-W-QP-JAN10.PDF) because that's probably easier than trying to interpret my notation.

I've done part (a) and got dy/dx= 4(1+tan^2(4x)), I just have no clue how to do the next part.
Any help is appreciated.

Reply 1

SherlockHolmes

13

Original post by SweatyGoldfish

Could anyone help with this question (7b). I'll link it here (http://filestore.aqa.org.uk/subjects/AQA-MPC3-W-QP-JAN10.PDF) because that's probably easier than trying to interpret my notation.

I've done part (a) and got dy/dx= 4(1+tan^2(4x)), I just have no clue how to do the next part.
Any help is appreciated.

Write 1+tan^2(4x) as sec^2(4x). Now you can differentiate dy/dx using the chain rule.

Reply 2

O.huxley

Original post by SweatyGoldfish

Could anyone help with this question (7b). I'll link it here (http://filestore.aqa.org.uk/subjects/AQA-MPC3-W-QP-JAN10.PDF) because that's probably easier than trying to interpret my notation.

I've done part (a) and got dy/dx= 4(1+tan^2(4x)), I just have no clue how to do the next part.
Any help is appreciated.

You know that y = tan(4x)

So try using the chain rule on your answer for a, sticking with tan(4x) for the start, then sub in for y at the end.

You'll need to know the trigonometric identity for tan^2(x) and sec^2(x).

Reply 3

SweatyGoldfish

OP

1

Original post by SherlockHolmes

.

Original post by O.huxley

.

Having trouble differentiating sec4x, how would I do that?

EDIT: Never mind, figured that part out

(edited 9 years ago)

Reply 4

SweatyGoldfish

OP

1

Cheers guys, got the answer

Reply 5

O.huxley

You dont need to:

4(1 + tan^2(4x)) = 4 + 4tan^2(4x)

So differentiating gives 4 = 0, and you can differentiate 4tan^2(4x) using the chain rule :smile:

AQA JAN 2010 PC3 Question 7(b) help

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