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Math Help - Derivatives

  1. #1
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    Derivatives

    I just want someone to check my derivatives to see if I did them right:

    f(x) = \frac{(ax+b)}{(x-1)(x-4)}


    f'(x) = \frac{(ax+b)(2x-5)-a(x-4)(x-1)}{((x-1)(x-4))^2}


    f''(x) = \frac{(ax+b)}{-(2x-5)^3(x^2-5x+4)^2}
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  2. #2
    Senior Member topher0805's Avatar
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    This is assuming that a and b are constants?

    First multiply out the bottom.

    \frac {(ax+b)}{(x-1)(x-4)} = \frac {ax+b}{x^2-5x+4}

    The quotient rule states that:

    \frac {d}{dx}(\frac {u}{v})=\frac {vu'-uv'}{v^2}

    So, by the quotient rule we have:

    \frac {(x^2-5x+4)(ax+b)'-(ax+b)(x^2-5x+4)'}{(x^2-5x+4)^2}

    = \frac {(x^2-5x+4)(a)-(ax+b)(2x-5)}{(x^2-5x+4)^2}

    = \frac {a(x-4)(x-1)-(ax+b)(2x-5)}{(x-4)^2(x-1)^2}

    Seems like you understand it, but remember the quotient rule:

    (bottom*derivative of the top)-(top*derivative of the bottom) / (bottom squared)
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  3. #3
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    Quote Originally Posted by topher0805 View Post
    = \frac {a(x-4)(x-1)-(ax+b)(2x-5)}{(x-4)^2(x-1)^2}
    So that is the correct first derivative, right?
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  4. #4
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    ***I basically need the second derivative so I can find the variables a and b given that the relative minimum is (2,-1). (Unless there is another way to solve the problem WITHOUT the second derivative.)


    Sorry for the double post, but something doesn't seem right.

    Here's my second derivative (too big to put into fraction form, I guess):

    Numerator: [(x-1)^2(x-4)^2][2ax-5a-4ax-2b-5a] -[a(x-1)(x-4)-(ax+b)(2x-5)][(4x^3-66x) -(30x^2+40)

    Denominator:
    (x-1)^4(x-4)^4
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  5. #5
    Senior Member topher0805's Avatar
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    Ok, I was assuming that a and b were constants. If they are variables that changes everything.
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  6. #6
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    Quote Originally Posted by topher0805 View Post
    Ok, I was assuming that a and b were constants. If they are variables that changes everything.
    No, they're constants. (Sorry about that. )
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  7. #7
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    Quote Originally Posted by Cursed View Post
    ***I basically need the second derivative so I can find the variables a and b given that the relative minimum is (2,-1). (Unless there is another way to solve the problem WITHOUT the second derivative.)

    [snip]
    The point (2, -1) is on the curve.

    f(2) = -1:

    2 = 2a + b .... (1)

    f'(2) = 0:

    b = 0 .... (2)

    Solve (1) and (2) simultaneously.
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  8. #8
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    So if I did this correctly, my answers of a=-1 and b=4 are correct, right?
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  9. #9
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    Quote Originally Posted by Cursed View Post
    So if I did this correctly, my answers of a=-1 and b=4 are correct, right?
    *Ahem* Take another look at equation (2) .......
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  10. #10
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    Quote Originally Posted by mr fantastic View Post
    *Ahem* Take another look at equation (2) .......
    And you get equation (2) as follows:

    0 = a(2 - 4)(2 - 1) - (2a + b)(4 - 5)

    => 0 = a(-2)(1) - (2a + b) (-1)

    => 0 = -2a + 2a - b

    => b = 0.
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