The Student Room Group

Differentiation help!! A2

Screenshot 2016-10-25 at 20.21.05.png SO i need help on the part (iii)
So far i've got
Unparseable latex formula:

f(x-1)= \frac{{e^{2x-1}} \times {e^{2x}}} {{x^2}\times{x-1}^2}}}

Reply 1
Original post by TheAlphaParticle
Screenshot 2016-10-25 at 20.21.05.png SO i need help on the part (iii)
So far i've got
Unparseable latex formula:

f(x-1)= \frac{{e^{2x-1}} \times {e^{2x}}} {{x^2}\times{x-1}^2}}}



I have no idea how you managed to get that for f(x-1). To get f(x-1) you should have subbed (x-1) into wherever you saw an (x) in the original f(x).
Reply 2
It should be f(x1)=e2(x1)(x1)2f(x-1) = \frac{e^{2(x-1)}}{(x-1)^2}
Reply 3
Is it k=e^-2 ?
Original post by jamestg
Is it k=e^-2 ?


Yes.
Reply 5
Original post by NotNotBatman
Yes.


Cracking. Well, that's my c3 revision done for today.
Reply 6
Original post by jamestg
Is it k=e^-2 ?


Questions like these never come up in Edexcel, they always ask some basic stuff like differentiate sin x :frown:

think I may do mei papers from now on

Spoiler

Reply 7
Original post by Naruke
Questions like these never come up in Edexcel, they always ask some basic stuff like differentiate sin x :frown:

think I may do mei papers from now on

Spoiler



Is MEI harder/more involved than edexcel? I'm currently on one past paper per week but I want to do more!
Reply 8
Original post by jamestg
Is MEI harder/more involved than edexcel? I'm currently on one past paper per week but I want to do more!


Well, there is a reason why Edexcel is the most widely taken exam board for maths :biggrin:
Reply 9
Original post by Naruke
Well, there is a reason why Edexcel is the most widely taken exam board for maths :biggrin:


That can't be the reason! It's probably down to resources :wink:

But I'll give a few papers a go :biggrin:
Original post by Naruke
Questions like these never come up in Edexcel, they always ask some basic stuff like differentiate sin x :frown:

think I may do mei papers from now on

Spoiler



If those exam question's aren't feeling challenging, you can have a go at deriving the general result for the nnth derivative of a product of two functions, fgfg. And who knows, perhaps this result can help you with whatever else. :smile:
Reply 11
Original post by RDKGames
If those exam question's aren't feeling challenging, you can have a go at deriving the general result for the nnth derivative of a product of two functions, fgfg. And who knows, perhaps this result can help you with whatever else. :smile:



let y=fg y = fg

y=fg+fg y' = f'g + fg'

y=fg+2fg+fg y'' = f''g+2f'g' + fg''

y=fg+3fg+3fg+fg y''' = f'''g +3f''g' + 3f'g'' + fg'''
.
.
.
Unparseable latex formula:

y^n = \displaystyle\sum_{r=0}^{n} \frac{n!}{r!(n-r)!} f^ng^n^-^r[br]

(edited 7 years ago)
Original post by Naruke
let y=fg y = fg

y=fg+fg y' = f'g + fg'

y=fg+2fg+fg y'' = f''g+2f'g' + fg''

y=fg+3fg+3fg+fg y''' = f'''g +3f''g' + 3f'g'' + fg'''
.
.
.
Unparseable latex formula:

y^n = f^ng + n(f^n^-^1g^n^-^2+f^n^-^2g^n^-^1)+fg^n



Does that hold for n=4?
Reply 13
Original post by RDKGames
Does that hold for n=4?


nope :colonhash: oh i see now

pascal
(edited 7 years ago)

Quick Reply

Latest