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    Name:  Screenshot 2016-10-25 at 20.21.05.png
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Size:  23.2 KB SO i need help on the part (iii)
    So far i've got  f(x-1)= \frac{{e^{2x-1}} \times {e^{2x}}} {{x^2}\times{x-1}^2}}}
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    (Original post by TheAlphaParticle)
    Name:  Screenshot 2016-10-25 at 20.21.05.png
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Size:  23.2 KB SO i need help on the part (iii)
    So far i've got  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).
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    It should be f(x-1) = \frac{e^{2(x-1)}}{(x-1)^2}
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    Is it k=e^-2 ?
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    (Original post by jamestg)
    Is it k=e^-2 ?
    Yes.
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    (Original post by NotNotBatman)
    Yes.
    Cracking. Well, that's my c3 revision done for today.
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    (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

    think I may do mei papers from now on
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    Show

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

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

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

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

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

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

    k= e^2^(^x^-^1^)^-^(^2^x^)

    k = e^-^2
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    (Original post by Naruke)
    Questions like these never come up in Edexcel, they always ask some basic stuff like differentiate sin x

    think I may do mei papers from now on
    Spoiler:
    Show

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

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

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

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

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

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

    k= e^2^(^x^-^1^)^-^(^2^x^)

    k = e^-^2
    Is MEI harder/more involved than edexcel? I'm currently on one past paper per week but I want to do more!
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    (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
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    (Original post by Naruke)
    Well, there is a reason why Edexcel is the most widely taken exam board for maths
    That can't be the reason! It's probably down to resources

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

    think I may do mei papers from now on
    Spoiler:
    Show

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

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

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

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

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

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

    k= e^2^(^x^-^1^)^-^(^2^x^)

    k = e^-^2
    If those exam question's aren't feeling challenging, you can have a go at deriving the general result for the nth derivative of a product of two functions, fg. And who knows, perhaps this result can help you with whatever else.
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    (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 nth derivative of a product of two functions, fg. And who knows, perhaps this result can help you with whatever else.

    let  y = fg

     y' = f'g + fg'

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

     y''' = f'''g +3f''g' + 3f'g'' + fg'''
    .
    .
    .
     y^n = \displaystyle\sum_{r=0}^{n} \frac{n!}{r!(n-r)!} f^ng^n^-^r
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    (Original post by Naruke)
    let  y = fg

     y' = f'g + fg'

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

     y''' = f'''g +3f''g' + 3f'g'' + fg'''
    .
    .
    .
     y^n = f^ng + n(f^n^-^1g^n^-^2+f^n^-^2g^n^-^1)+fg^n
    Does that hold for n=4?
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    (Original post by RDKGames)
    Does that hold for n=4?
    nope oh i see now

    pascal
 
 
 
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