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Math Help - Question on canceling fractional quantities

  1. #1
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    Question on canceling fractional quantities

    This is more of an algebra question than calculus, but it involves derivatives so I thought I'd post it here.

    I have  f(x) = \frac{5x}{(3-5x)^2}

    When I apply the quotient rule I have:  f'(x) = \frac{5x[2(3-5x)]*(-5)-5(3-5x)^2}{[(3-5x)^2]^2} or  f'(x) = \frac{-50x(3-5x)-5(3-5x)^2}{(3-5x)^4} simplified

    My question is can I take the quantity  (3-5x)^4 in the denominator and simplify it with the same quantities in the numerator? Since I am essentially subtracting 2 quantities of  (3-5x) , would I be cancelling out the term to the far right completely, and leaving the term on the far left with a quantity of -1? Thus that leaves the denominator with the 3 quantities? I remember my professor saying something about all derivatives of rational functions will increment by 1 in the denominator.

    To get:  f'(x) = \frac{-50x-5}{(3-5x)^3}

    This is just something I never fully understood with quantities learning algebra, but now I kind of have to face it with these kinds of calculus problems.
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  2. #2
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    Re: Question on canceling fractional quantities

    FYI - Your professors comment was about what happens *WHILE* taking a derivative, not while algebraically simplifying a derivative you've already taken.

    When you cancel common factors in fractions algebraically, you aren't doing anything different than cancelling common factors arithmetically.

    As an aside, if you're ever unsure of some algebra rule, try thinking about a parallel arithmetic problem. It might prove enlightening. Remember that the algebra stuff only works because, hidden behind all those variables, it's all just numbers, so you're doing nothing different than arithmetic in disguise (well, until you get to calculus...).

    You got to here (correctly it looked to me):  f'(x) = \frac{-50x(3-5x)-5(3-5x)^2}{(3-5x)^4}.

    You want to know if you can cancel *all* the (3-5x) bits in the numerator, giving this:  ?= \frac{-50x-5}{(3-5x)^3}?

    The answer is... no. For instance, plug in x=1 and you can see that those two are not equal.

    In arithmetic, could you write \frac{2(3)-7(3)^2}{3^4} \ ? \ = \ ? \ \frac{2-7}{3^3}. The answer is no, that's incorrect.

    The way to simplify fractions using cancellation is to make sure that the factors you're cancelling multiply the ENTIRE rest of the numerator and denominator. Like this:

    \frac{2(3)-7(3)^2}{3^4} = \frac{(3)\left(2-7(3)^1\right)}{(3)(3^3)} = \frac{(3)}{(3)} \frac{\left(2-7(3)^1\right)}{(3^3)} = (1) \frac{\left(2-7(3)^1\right)}{(3^3)}

    = \frac{2-7(3)}{3^3}


    In your case:

    f'(x) = \frac{-50x(3-5x)-5(3-5x)^2}{(3-5x)^4} = \frac{(3-5x)\left(-50x-5(3-5x)^1\right)}{(3-5x)^4}

    = \frac{(3-5x)}{(3-5x)} \ \frac{\left(-50x-5(3-5x)^1\right)}{(3-5x)^3} = \frac{-50x-5(3-5x)^1}{(3-5x)^3}.

    And then simplify more: f'(x) = \frac{-50x-5(3-5x)^1}{(3-5x)^3} = \frac{-50x +(25x-15)}{(3-5x)^3}= \frac{-25x -15}{(3-5x)^3}.

    Thus f'(x) = (-5)\frac{5x +3}{(3-5x)^3} = (5)\frac{5x +3}{(5x-3)^3}.
    Last edited by johnsomeone; October 3rd 2012 at 10:34 AM.
    Thanks from AZach
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  3. #3
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    Re: Question on canceling fractional quantities

    It looks like he had a very circumlocutory way of saying that you cannot divide when there is addiction or subtraction within the fraction. (If it's inside of parenthesis, it doesn't count.)
    Last edited by Nervous; October 3rd 2012 at 01:44 PM.
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  4. #4
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    Re: Question on canceling fractional quantities

    I think I get it now. I think I just got into a bad habit of canceling in rational expressions, but I see now that the correct way to look at is to almost imagine that there's a *1 on the side of the expression, and if a part of the expression can get simplified then that part must be also equivalent to *1. I really appreciate the help, thanks!!

    If somebody has a chance, can they check my math for this Q: find equation of tangent line @ x = 2  f(x) =\frac{(5x)}{(3-5x)^5}  f(2) = \frac {(5(2))}{(3-5(2))^5} = -0.0636943 So, the point where the tangent line touches f(x) is (2, -0.0636943)

    I got  f'(x) = \frac{-50x-5(3-5x)}{(3-5x)^6}

     f'(2) = \frac{(-50(2)-5(3-5(2))}{(3-5(2))^6} = -0.00055249

    Point slope form got me:  y + 0.0636943 = -0.00055249(x-2) and simplified  y = -0.00055249x-0.625893

    Do my steps look about right?
    Last edited by AZach; October 3rd 2012 at 09:54 PM.
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