differentiation

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    given that y = x^2/ ax+b (a, b doesnt equal 0) show tat d^2y/dx^2 = 2b^2/(ax+b)^3

    dy/dx= ax^2 + 2bx / (ax+b)^2

    where do i go from here
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    (Original post by Custardcream000)
    given that y = x^2/ ax+b (a, b doesnt equal 0) show tat d^2y/dx^2 = 2b^2/(ax+b)^3

    dy/dx= ax^2 + 2bx / (ax+b)^2

    where do i go from here
    Assuming it's correct (I can't be bothered to check) then just differentiate it again.
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    (Original post by RDKGames)
    Assuming it's correct (I can't be bothered to check) then just differentiate it again.
    so d^2y/dx^2 = [(ax+b)^2(2ax+2b) - (ax^2 +2bx)(2a(ax+b))]/ (ax+b)^4
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    (Original post by Custardcream000)
    so d^2y/dx^2 = (ax+b)^2(2ax+2b) - (ax^2 +2bx)(2a)/ (ax+b)^4
    Nope. Check it again and post your working if you get stuck.
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    (Original post by RDKGames)
    Nope. Check it again and post your working if you get stuck.
    d^2y/dx^2 = (ax+b)^2(2ax+2b) - (ax^2 +2bx)(2a(ax+b))/ (ax+b)^4
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    (Original post by Custardcream000)
    d^2y/dx^2 = (ax+b)^2(2ax+2b) - (ax^2 +2bx)(2a(ax+b))/ (ax+b)^4
    No...

    You shouldn't even get a power of 4.
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    (Original post by Custardcream000)
    d^2y/dx^2 = (ax+b)^2(2ax+2b) - (ax^2 +2bx)(2a(ax+b))/ (ax+b)^4
    It might be easier to follow if you work it out on paper and then post a picture of your working.
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    (Original post by RDKGames)
    No...

    You shouldn't even get a power of 4.
    i'm using the (vu' - uv')/v^2 method though
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    please put brackets around the top of the fraction.
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    (Original post by Custardcream000)
    i'm using the vu' - uv'/v^2 method though
    In which case one power should cancel and it's hard to see where without your working.

    u=ax^2+2bx \Rightarrow u'=2ax+2b

    v=(ax+b)^2 \Rightarrow v'=2a(ax+b)

    \displaystyle \frac{vu'-uv'}{v^2}=\frac{(2ax+2b)(ax+b)^2-(ax^2+bx)[2a(ax+b)]}{(ax+b)^4}

    You can factor out one ax+b from the numerator straight away and cancel a power. The rest is just tidying up via expansion and cancellation.

    Please use latex or square brackets next time when writing out your fractions, or show your working, I got confused when looking at them.
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    (Original post by RDKGames)
    In which case one power should cancel and it's hard to see where without your working.

    u=ax^2+2bx \Rightarrow u'=2ax+2b

    v=(ax+b)^2 \Rightarrow v'=2a(ax+b)

    \displaystyle \frac{vu'-uv'}{v^2}=\frac{(2ax+2b)(ax+b)^2-(ax^2+bx)[2a(ax+b)]}{(ax+b)^4}

    You can factor out one ax+b from the numerator straight away and cancel a power. The rest is just tidying up via expansion and cancellation.

    Please use latex or square brackets next time when writing out your fractions, or show your working, I got confused when looking at them.
    thanks so i get

    {2a^3x^2 + 4a^2bx^2 + 6ab^2x +2b^3 - 2a^2x^2 + 2abx} / (ax+b)^3
 
 
 
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