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Edexcel C3,C4 June 2013 Thread

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Original post by Chris-69
Hey,

could anyone explain how the derivative of y = ln10(log10x) = 1/x


Thanks :smile: !


Anyone know how differentiate it?
Could someone help we this?
If you integrate (1+2x)^-1, is the answer 0.5 ln (1+2x) or 0.5 ln I1+2xI (with a modulus)?????

Thanks x
Reply 4522
Avatar for pjanoo
pjanoo
2
Original post by kuku2013
am I right I thinking when you differentiate you multiply by ln(2) and when you integrate you divide by ln(2)


That's how I do it :smile:
Reply 4523
Avatar for KalSA
KalSA
8
Original post by monkey101
Could someone help we this?
If you integrate (1+2x)^-1, is the answer 0.5 ln (1+2x) or 0.5 ln I1+2xI (with a modulus)?????

Thanks x


I usually put mine with a modulus although marking schemes put brackets.. go with modulus though.
Original post by Story
Yes, but how would it split all the way?


Zaphod has explained it in an above post :smile:
After C3 tested alot of knowledge of earlier modules spanning from GCSE what sort of things do people recomend look at? - Things from the earlier modules/gcse which could be relevant to the C4 topics.

Thanks
Reply 4526
Avatar for Jullith
Jullith
Original post by wndms
Can anyone please help me with this c4?
Fir part c) I know you have to equate dy/dx to 0. But then what do you do after? The answer is supposed to be (pi,-1). I don't understand .. thanks!

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Hope this helps:

C4 5.png
Reply 4527
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fayled
9
Original post by Chris-69
Anyone know how differentiate it?


Change of base formula.

log10x=lnx/ln10
Original post by kuku2013
am I right I thinking when you differentiate you multiply by ln(2) and when you integrate you divide by ln(2)


For 2x, yes.

In general,

ddx(ax)=axln⁡a\dfrac{d}{dx}(a^x) = a^x \ln a

∫ax dx=axln⁡a+C\displaystyle \int a^x \ dx = \dfrac{a^x}{\ln a} + C
Reply 4529
Avatar for Story
Story
Original post by Zaphod77
Ok, there are 2 methods you could use. 1 is that you make the top u+4-4, so then you could split the fractions into (u+4)/(u+4) - 4/(u+4), which you can then integrate, as the first fraction simplifies down to equal 1. The other method is dividing u by u+4, which gives you a whole number and a remainder which you can then integrate :smile:


Thank you. If you integrate it would it be ...u - 4ln(u+4) or can you not do that?
Does anyone have any good tips about connected rate of change.... Im really bad at it :frown:

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Original post by Nilsdejongh
After C3 tested alot of knowledge of earlier modules spanning from GCSE what sort of things do people recomend look at? - Things from the earlier modules/gcse which could be relevant to the C4 topics.

Thanks


After C3, I started revising M2 and FP3, just to be safe.
Original post by Story
Thank you. If you integrate it would it be ...u - 4ln(u+4) or can you not do that?


That's correct :smile: -4 is a scalar, so that's fine!
Reply 4533
Avatar for Jullith
Jullith
Still need help with part c of this:

C4 4.png

And part c of this:
(I'm confused as to what the limits are)

C4 3.JPG

Help would be appreciated, thanks!
Reply 4534
Avatar for Story
Story
Original post by Zaphod77
That's correct :smile: -4 is a scalar, so that's fine!


Thank you so much! You are really helpful!
Original post by Story
Thank you so much! You are really helpful!


No problem, happy to help :smile:
Reply 4536
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wndms
1
Original post by Jullith
Hope this helps:

C4 5.png


THANK YOU

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Reply 4537
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KD35
15
Lets set ourselves some hard questions guys!

Integrate 4^(2x^2)

Thats 4 to the power 2x^2
does anyone have a list of formulas we need to know?
Original post by KD35
Lets set ourselves some hard questions guys!

Integrate 4^(2x^2)

Thats 4 to the power 2x^2


Is it just the x squared, or the 2x all squared?

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