[skibur]

(Original post by Horchata)
It's to find the 'valid for' range for maclaurin. I'm guessing we don't need to know it..

Can anyone advise me on how to find th maclaurin expansion for tanx? I find that really difficult. I would be so happy if we had to expand cos or sin!

differential of tanx is sec^2x, differential of that 2sec^2xtanx i think, doubt they'll ask for anymore than that.

The first diffferential I knew from C3 and then for the second one I do(Secx)^2 then differentiate it normaly to 2secx then multiply my the differential of secx which is secxtanx.

reading this over it makes no sense, sorry but its hard to explain over text

[Horchata]

(Original post by skibur)
differential of tanx is sec^2x, differential of that 2sec^2xtanx i think, doubt they'll ask for anymore than that.

The first diffferential I knew from C3 and then for the second one I do(Secx)^2 then differentiate it normaly to 2secx then multiply my the differential of secx which is secxtanx.

reading this over it makes no sense, sorry but its hard to explain over text

I understand what you mean lol. Do you happen to know the next one? thanks btw

[reaperguy]

(Original post by Horchata)
I understand what you mean lol. Do you happen to know the next one? thanks btw

no there was a question asking for a Macularen Series for tanx upto x^4. You just have to use product rule and chain rule again and again
so for 2sec^2xtanx
uv'+vu'=dy/dx

where u=2sec^2x
and v=tanx

u'=chain rule so you can start with a new equation
for example t=2sec^2x
let S=Secx so t=2S^2
ds/dx=secxtanx
and dt/ds=4S
so dt/dx which is u' is 4Sec^2xtanx

So uv'+vu'= (2Sec^2x)(sec^2x)+(Tanx)(4Sec^2x tanx)

since you put in f(0) tan(0)=0
so you have 2Sec^4(0)

I hope that this is clear lol

[SophieArmstrong]

(Original post by Horchata)
Don't worry my teacher just sent it to us lol what a troll. Thanks for your effort though

Please could you upload the mark scheme to here?

[MHRed]

(Original post by Horchata)
It's to find the 'valid for' range for maclaurin. I'm guessing we don't need to know it..

Nah, don't think so (I think it normally tells you in the formulae booklets), and if not, you can normally adapt it from the info given there

[skibur]

found january 12 paper actually quite difficult

[Horchata]

(Original post by reaperguy)
no there was a question asking for a Macularen Series for tanx upto x^4. You just have to use product rule and chain rule again and again
so for 2sec^2xtanx
uv'+vu'=dy/dx

where u=2sec^2x
and v=tanx

u'=chain rule so you can start with a new equation
for example t=2sec^2x
let S=Secx so t=2S^2
ds/dx=secxtanx
and dt/ds=4S
so dt/dx which is u' is 4Sec^2xtanx

So uv'+vu'= (2Sec^2x)(sec^2x)+(Tanx)(4Sec^2x tanx)

since you put in f(0) tan(0)=0
so you have 2Sec^4(0)

I hope that this is clear lol

Omg thank you! I found another method on youtube http://www.youtube.com/watch?v=43QNY-edoys

not sure if we can use it though

[Horchata]

Here is Jan 2012 with the mark scheme in the same file

[skibur]

its in the afternoon right...

[albus]

Has anyone got a way to remember hyperbolic identities.
The one for cosh(2x), sinh(2x),
and its cosh^2-sinh^2=1 and cosh^2+sinh^2=cosh(2x), am I right.

[Reminisce]

(Original post by albus)
Has anyone got a way to remember hyperbolic identities.
The one for cosh(2x), sinh(2x),
and its cosh^2-sinh^2=1 and cosh^2+sinh^2=cosh(2x), am I right.

Yeah just learn these then you can derive the rest.

[Petro_99]

(Original post by skibur)
its in the afternoon right...

yes

[SophieArmstrong]

(Original post by albus)
Has anyone got a way to remember hyperbolic identities.
The one for cosh(2x), sinh(2x),
and its cosh^2-sinh^2=1 and cosh^2+sinh^2=cosh(2x), am I right.

These ones are all in the formula book and you can derive any others you might need from these So you don't have to learn any!
cosh^2 x - sinh^2 x = 1
sinh 2x = 2sinh x cosh x
cosh 2x = cosh^2 x + sinh^2 x

[albus]

(Original post by SophieArmstrong)
These ones are all in the formula book and you can derive any others you might need from these So you don't have to learn any!
cosh^2 x - sinh^2 x = 1
sinh 2x = 2sinh x cosh x
cosh 2x = cosh^2 x + sinh^2 x

What page is it in the formula book.
I never knew it was in there.

[SophieArmstrong]

(Original post by albus)
What page is it in the formula book.
I never knew it was in there.

Page 2 It's a little section below all the infinite series stuff!

[skibur]

Has anyone done the Jan12 paper, how did everyone find it?

[skibur]

(Original post by albus)
What page is it in the formula book.
I never knew it was in there.

How the **** did you know what 6 marker was going to be on the physics nuclear paper

[albus]

(Original post by SophieArmstrong)
Page 2 It's a little section below all the infinite series stuff!

Thanks for that.
Is there any more useful FP2 stuff in the formula book.

[albus]

(Original post by skibur)
How the **** did you know what 6 marker was going to be on the physics nuclear paper

I didn't know what was coming up on the paper, I had an intelligent guess.

[skibur]

(Original post by albus)
I didn't know what was coming up on the paper, I had an intelligent guess.

hmmm let you off then... Im sure you did it for unit 4 too