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# Edexcel C4 Revision Thread (28/06 14:00) watch

1. (Original post by nas7232)
you should differentiate it via the product rule.
Product & Chain Rule at the same time
2. (Original post by TheWolf)

naz i do use the product rule
ah sorry i thought you said you used the chain rule
3. (Original post by samdavyson)
Product & Chain Rule at the same time
ah
4. (-5ln(2/3) - 2ln(5/3)) - (-5ln(5/6) - 2ln(4/3))

=> -5ln(2/3) + 5ln(5/6) - 2ln(5/3) + 2ln(4/3)

=> 5ln([5/6]/[2/3]) + 2ln([4/3]/[5/3])

=> 5ln(5/4) + 2ln(4/5)

Argh! The question disappeared!
5. Thanks. **** i totally missed that ='(
6. Good Luck All.

Good night.
7. (Original post by samdavyson)
Good Luck All.

Good night.
Good luck to you also!
8. Q2 - Mock paper.

Ok, so when you get to:

3x²-2x dy/dx + 2y - 4 + 3y² dy/dx = 0

Which way do you rearrange that?
9. Hello all, i've been doing some questions from the Edexcel Papers of the Mathmo gmail account...got stuck on this question:

int 1/(1-x²)^³/² dx

=

X/(1-x²)^½ + C

Using X=sinθ, show....etc

In the marksceme they have used:

int cosθ/cos³θ = int sec²...

And so on, but i am unsure of how they got there. Any hints?

10. (Original post by bavman)
Hello all, i've been doing some questions from the Edexcel Papers of the Mathmo gmail account...got stuck on this question:

int 1/(1-x²)^³/² dx

=

X/(1-x²)^½ + C

Using X=sinθ, show....etc

In the marksceme they have used:

int cosθ/cos³θ = int sec²...

And so on, but i am unsure of how they got there. Any hints?

int 1/(1-x²)^³/² dx
x = sinθ
dx/dθ = cos θ
dx = cosθ dθ

(1-x²)^3/2 = (1-sin²θ )^3/2 = (cos²θ )^3/2 = cos³θ

∫ 1/cos³θ x cosθ dθ
∫ cosθ/cos³θ dθ
∫ sec²θ dθ

and take it away from there!
11. isnt C4 at 1pm
12. Depends. Anything from 1 to 2, depending on the centre.
13. Specimen 7b =0 how to di it?
14. (Original post by TheWolf)
Specimen 7b =0 how to di it?
look at this thread dealing with the exact same question:
http://www.thestudentroom.co.uk/t126670.html
15. How to approach part a?
Attached Images

16. I'd say find the equation of the line AB in the form r = a +λb, where a = (9i-8j+2k) and b = (4j+k) - (9i-8j+2k).
Then equate the two line equations, and equate coefficients of i, j and k, giving 3 simultaneous equations. Finally, solve, and sub either λ or µ back into one of the line equations giving the position vector of C (i.e. the point where the two roads intersect).
17. Cheers!. :3
18. say if i found the gradient to be 2cos2t/2sint and I wanted to find what the gradient was at pi over 3, how would I type this into my calculator, starting with the top first? would be 2 times pi over 3, cos the answer and then times this answer by 2 for the top row?
say if i found the gradient to be 2cos2t/2sint and I wanted to find what the gradient was at pi over 3, how would I type this into my calculator, starting with the top first? would be 2 times pi over 3, cos the answer and then times this answer by 2 for the top row?
That sounds okay.
You could alternately abuse brackets and do (2cos(2pi/3))/(2sin(2pi/3)) but a calculator isn't essential has cos(2pi/3)=-1/2 and sin(pi/3)=rt3/2 IIRC
say if i found the gradient to be 2cos2t/2sint and I wanted to find what the gradient was at pi over 3, how would I type this into my calculator, starting with the top first? would be 2 times pi over 3, cos the answer and then times this answer by 2 for the top row?
i would say, just dont use a calculator!
cos (2pi/3) = -cos(pi/3) = -0.5
sin (pi/3) = √3/2
then just plug it in!

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