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Math Help - simplifying a quotient rule question of f"

  1. #1
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    simplifying a quotient rule question of f"

    Find f '(x) and f ''(x) of f(x)= (x^2)/(1+4x)

    I found f':
    f'= (2x+4x^2)/(1+4x)^2.

    for f"
    [(2+8x)(1+4x)^2 - 2(1+4x)(4)]/[(1+4x)^2]^2

    *I know this is correct, but I have to input it in an online grader so it has to be in it's most simplified form.*
    so...
    factoring a (1+4x)^2=
    (1+4x)[(2+8x)(1+4x)-8]/[(1+4x)^4]
    (1+4x)[(2+8x+8x+32x^2)-8]/[(1+4x)^4]
    (1+4x)[(2+8x+8x+32x^2)-8/[(1+4x)^4]
    (1+4x)[(16x+32x^2-6/[(1+4x)^4]
    factoring out a 2:
    (2+8x)[(16x^2+8x-3)]/[(1+4x)^4] *was marked wrong*

    Any arithmetic/calculus errors or can it be simplified further?
    Last edited by hazecraze; October 8th 2009 at 08:43 AM.
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  2. #2
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by hazecraze View Post
    Find f '(x) and f ''(x) of f(x)= (x^2)/(1+4x)

    I found f':
    f'= (2x+4x^2)/(1+4x)^2.

    for f"
    [(2+8x)(1+4x)^2 - 2(1+4x)(4)]/[(1+4x)^2]^2

    *I know this is correct, but I have to input it in an online grader so it has to be in it's most simplified form.*
    so...
    factoring a (1+4x)^2=
    (1+4x)[(2+8x)(1+4x)-8]/[(1+4x)^4]
    (1+4x)[(2+8x+8x+32x^2)-8]/[(1+4x)^4]
    (1+4x)[(2+8x+8x+32x^2)-8]/[(1+4x)^4]
    (1+4x)[(16x+32x^2-6]/[(1+4x)^4]
    factoring out a 2:
    (2+8x)[(16x^2+8x-3)]/[(1+4x)^4] *was marked wrong*

    Any arithmetic/calculus errors or can it be simplified further?
    what's \frac {2+8x} {1 + 4x}?
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  3. #3
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    2, so 2(16x^2+8x-3)/(1+4x)^3?
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  4. #4
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    Have you checked to make sure that 16x^2+ 8x- 3 doesn't have a factor of 4x+ 1?
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  5. #5
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by hazecraze View Post
    2, so 2(16x^2+8x-3)/(1+4x)^3?
    You can also factorise the quadratic. Go on, give it a go.
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    Have you checked to make sure that 16x^2+ 8x- 3 doesn't have a factor of 4x+ 1?
    Shouldn't it be 4x- 1? (4x- 1)(4x+3)= 16x^2+ 12x-4x- 3= 16x^2+ 8x- 3. That messes up the canceling of the denominator though, which I can see now is where we were going.
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  7. #7
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by hazecraze View Post
    Find f '(x) and f ''(x) of f(x)= (x^2)/(1+4x)

    I found f':
    f'= (2x+4x^2)/(1+4x)^2.

    for f"
    [(2+8x)(1+4x)^2 - 2(1+4x)(4)]/[(1+4x)^2]^2

    *I know this is correct, but I have to input it in an online grader so it has to be in it's most simplified form.*
    I'm not sure it is.

    Let's put math delimiters round it to start with so we can see the silly thing:

    f(x)= \frac {x^2} {1+4x}

    f'(x) = \frac {2x+4x^2} {(1+4x)^2}, yep, I'll agree there ...

    Then you got your quotient rule d \left({\frac u v}\right) = \frac {v du - u dv} {v^2}

    u = 2x+4x^2, v = (1+4x)^2 so du = 2 + 8x, dv = 2 (1+4x) 4 = 8 (1+4x).

    So f''(x) = \frac {(1+4x)^2 (2 + 8x) - 8 (2x+4x^2) (1+4x)} {(1+4x)^4}

    I'll leave you to simplify it, I've got things to do today.
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  8. #8
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    Doh! Don't know how I missed that. So, \frac {2} {1+4x^3}. I figured they'd give us something easy to simplify.
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  9. #9
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by hazecraze View Post
    Doh! Don't know how I missed that. So, \frac {2} {1+4x^3}. I figured they'd give us something easy to simplify.
    You're not the only one to miss it - we all did! (to start with, at least.)
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