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Math Help - Derivative question using the Quotient rule

  1. #1
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    Derivative question using the Quotient rule

    Ithe problem is as follows: d/dx[1-5x/1+5x] I need to find f'(x) and f''(x)
    I found f'(x) but the second derivative is giving me problems. Here is what I did for both: d/dx[1-5x/1+5x] = (1+5x) d/dx [1-5x] - (1-5x) d/dx [1+5x] and all this is divided by (1+5x)^2
    = (1+5x)(-5) - (1-5x)(5)/(1+5x)^2 =
    -5 -25x - 5 + 25x/(1+5x)^2 =
    -10/(1-5x) Now I tried to get f''(x) unsuccessfully I might add.
    (1+10x+25x^2) d/dx [-10] - (-10) d/dx [1+10x+25x^2] all this is divided by
    (1+10x+25x^2)^2 =
    1+10x+25x^2 - (-10)(10)(50x)/(1+10x+25x^2)^2 =
    1+5010x+25x^2/(1+10x+25x^2)^2

    Please let me know where I went off track.
    THANK YOU!
    Keith
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  2. #2
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    Quote Originally Posted by kcsteven View Post
    Ithe problem is as follows: d/dx[1-5x/1+5x] I need to find f'(x) and f''(x)
    I found f'(x) but the second derivative is giving me problems. Here is what I did for both: d/dx[1-5x/1+5x] = (1+5x) d/dx [1-5x] - (1-5x) d/dx [1+5x] and all this is divided by (1+5x)^2
    = (1+5x)(-5) - (1-5x)(5)/(1+5x)^2 =
    -5 -25x - 5 + 25x/(1+5x)^2
    Hello, Keith,

    all calculations are correct but the next step:
    = -10/(1+5x)^2

    To get f''(x) it would be better to write f'(x) = (-10)*(1+5x)^(-2)
    Now you only have to use the chain rule:
    f''(x) = (-10)*(-2)*(1+5x)^(-3)*(5) = (100)/((1+5x)^3)

    EB
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  3. #3
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    Quote Originally Posted by kcsteven View Post
    Ithe problem is as follows: d/dx[1-5x/1+5x] I need to find f'(x) and f''(x)
    I found f'(x) but the second derivative is giving me problems. Here is what I did for both: d/dx[1-5x/1+5x] = (1+5x) d/dx [1-5x] - (1-5x) d/dx [1+5x] and all this is divided by (1+5x)^2
    = (1+5x)(-5) - (1-5x)(5)/(1+5x)^2 =
    -5 -25x - 5 + 25x/(1+5x)^2 =
    -10/(1-5x) Now I tried to get f''(x) unsuccessfully I might add.
    (1+10x+25x^2) d/dx [-10] - (-10) d/dx [1+10x+25x^2] all this is divided by
    (1+10x+25x^2)^2 =
    1+10x+25x^2 - (-10)(10)(50x)/(1+10x+25x^2)^2 =
    1+5010x+25x^2/(1+10x+25x^2)^2

    Please let me know where I went off track.
    THANK YOU!
    Keith

    Hello Keith!

    First, I know their was already a post submitted since I've been writing this, but I figured some teachers want you specifically to use the quotient rule over the power rule. Which sucks, because I love the power rule. So using only the quotient rule, your (small) mistakes are below:

    To begin, I, without looking at your work, equated f'(x) and got the same answer as you:
    f'(x) = -10/[(1+5x)^2]

    I think I've found your problem. When you are taking this part of your f''(x):

    (1+10x+25x^2) d/dx [-10]

    You say that the above argument equals (1+10x+25x^2). But d/dx of -10, or any constant, is 0, not 1. So:

    (1+10x+25x^2) d/dx [-10] equals 0 instead of (1+10x+25x^2)

    So looking at your work again:

    (1+10x+25x^2) d/dx [-10] - (-10) d/dx [1+10x+25x^2] all this is divided by
    (1+10x+25x^2)^2 =
    1+10x+25x^2 - (-10)(10)(50x)/(1+10x+25x^2)^2 =
    1+5010x+25x^2/(1+10x+25x^2)^2

    The result would change to:

    0 - (-10)(10)(50x)/(1+10x+25x^2)^2 =
    5000x/(1+10x+25x^2)^2

    The other problem lies at this point in your work:
    (-10) d/dx [1+10x+25x^2] = (-10)(10)(50x)
    (-10) d/dx [1+10x+25x^2] should equal:
    (-10) * (50x+10) which equals: -500x -100

    So we now have:

    0 - (-500x - 100) / (1+10x+25x^2)^2=
    (500x + 100) / (1+10x+25x^2)^2 =
    (500x + 100) / (1+5x)^4

    Now, if you really want to take it further,
    (1 +5x)^4 can be expanded to:
    625x^4 + 500x^3 + 150x^2 + 20x + 1

    So an optional f''(x) answer is:
    (500x+100)/(625x^4 + 500x^3 + 150x^2 + 20x + 1)
    But if it were me I'd leave the denominator unexpanded.

    So the final answers:
    f'(x) = (-10) / (1 + 5x)^2
    f''(x) = (500x + 100) / (1 + 5x)^4

    Hope this helps!

    P.S.

    My answer for f''(x) equals the answer from the above post:

    To get the above answer, we factor (100) out of the numerator:
    (500x +100) / (1 + 5x)^4 = (100)(1 + 5x) / (1 + 5x)^4
    The (1 + 5x)'s cancel out, giving us:
    100 / (1 + 5x)^3, the answer from the above post.
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  4. #4
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    Hello, Keith!

    f(x) .= .(1-5x)(1+5x) . Find f'(x) and f''(x)

    Your first derivative is correct: .f'(x) .= .10/(1 + 5x)
    . . but why multiply it out?


    EB had the best advice for the second derivative.

    You can do it head-on, but you must be more careful.


    . . . . . . . (1 + 5x)0 - 102(1 + 5x)5 . . . . -100(1 + 5x) . . . . . -100
    f''(x) . = . ---------------------------------- . = . --------------- . = . ------------
    . . . . . . . . . . . . (1 + 5x)^4 . . . . . . . . . . . .(1 + 5x)^4 . . . . .(1 + 5x)

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