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Differentiate This

  1. Apr 21, 2006 #1
    Find the derivative.

    http://img65.imageshack.us/img65/232/der429bj.gif [Broken]

    I really don't know which way to go with this.. It's just really complicated.
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Apr 21, 2006 #2


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    Well first ask yourself why it's complicated. Then ask yourself what you can do to get rid of that complication.
  4. Apr 21, 2006 #3
    well let me show you my work.. eh.. take me a year to type it out in mathtype :)..

    http://img232.imageshack.us/img232/6218/rawr8cx.gif [Broken]

    There's my work... *out of breath*

    I don't know how to simplify it any more. Or i just went in a wrong direction.
    Last edited by a moderator: May 2, 2017
  5. Apr 21, 2006 #4


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    First look at the second part of the equation. Do you notice that it's one function divided by another? You know how to find the derivative of each of those functions individually. Do you remember the quotient rule for derivatives?

    I guess another way to tackle this problem would just be to simplify the fraction.
    Last edited: Apr 21, 2006
  6. Apr 21, 2006 #5
    so, i'd take the derivative of 3x.. which would be 3... and the derivative of the second part using the quotient rule... so.. finding a common denominator isn't necessary?

    Bear with me... It's taking a lot of time.. i'm gonna post some work ... I'm just working on about 5 problems at once.
  7. Apr 21, 2006 #6


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    Right. The derivative of a product is the product of the derivatives.

    The quotient rule would work, yes. I guess you could combine the numberator into one fraction and try to simplify it.
  8. Apr 21, 2006 #7
    This is what I got.. by getting a common denominator in the numerator.. and making the numerator into one fraction.. then i flipped x-2 up.. multiplying the fraction by the reciprocal.. and that gave me 3x + [(x^2+4)/(x^2-x)]
    then i did the quotient rule and ended up getting...

    http://img82.imageshack.us/img82/6730/bleh2mm.gif [Broken]

    is this right, and is this simple enough?
    Last edited by a moderator: May 2, 2017
  9. Apr 22, 2006 #8
    Your application of the quotient rule is correct, but there are a couple of problems with your algebra.
    [tex]- \frac{\frac{2}{x}-\frac{3}{x-1}}{x-2} \neq \frac{x^2+4}{x^2-x} [/tex]
    [tex]- \left( \frac{2}{x}-\frac{3}{x-1} \right) = - \left( \frac{2(x-1)}{x(x-1)} - \frac{3(x)}{(x-1)(x)} \right) [/tex]

    Hopefully you can see where to go from here.
  10. May 11, 2006 #9
    Not quite -- Leibniz thought the same thing though for a while, I think, so you are in good company ;)

    But, the correct rule for differentiating a product of two functions is as follows:

    d{f(x)g(x)}/dx = (df(x)/dx)g(x) + f(x)(dg(x)/dx)


    (fg)' = f'g + fg'​

    Depending on which notation you like.
    Last edited: May 11, 2006
  11. May 12, 2006 #10


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    A partial fraction decomposition can make this complicated problem very simple.:smile:

    Let that complicated quotient be represented by y.

    You can see that the denominator of y is a cubic (x)(x-1)(x-2). So y should have the form A/x + B/(x-1) + C/(x-2). We now need to find A, B and C.

    Now let me introduce you to a very quick way to do that - it's called the "cover up method". This works when there are linear factors of the form (x - a) in the denominator. You simply cover up the (x - a) in the denominator and put x = a into the rest to get the numerator (A) of the A/(x-a) term.

    For example, let's work out A. Cover up the denominator x in the expression for y and put x = 0 in the rest of the expression. Instantly, you get A = 2/(-2) = -1.

    Similarly, you can easily get B (cover up the x-1 and put x = 1) and C (cover up x-2 and put x = 2) to be 3 and -2 respectively,

    (BTW, the reason this works has to do with multiplication. When you have an expression g(x) given by [tex]\frac{g(x)}{(x-a)(x-b)(x-c)} = \frac{A}{x-a} + \frac{B}{x-b} + \frac{C}{x-c}[/tex] and you multiply by one of the factors of the denominator (let's say (x-a)), you get [tex]\frac{g(x)}{(x-b)(x-c)} = A + \frac{B(x-a)}{x-b} + \frac{C(x-a)}{x-c}[/tex]. You can see that if you now let x = a, everything vanishes except the A on the right hand side and [tex]\frac{g(a)}{(a-b)(a-c)}[/tex] on the left hand side. So by evaluating that function on the left hand side, you've determined one coefficient very simply. Get it?)

    So we know that y = [tex]-\frac{1}{x} + \frac{3}{x-1} - \frac{2}{x-2}[/tex]

    and so the original function [tex]f(x) = 3x +\frac{1}{x} - \frac{3}{x-1} + \frac{2}{x-2}[/tex]

    which you can differentiate very easily by the Chain Rule (you don't need the Quotient Rule)

    [tex]f'(x) = 3 - \frac{1}{x^2} + \frac{3}{{(x-1)}^2} - \frac{2}{{(x-2)}^2}[/tex]

    I have to say that all this looks a little complicated, but when you get good at partial fraction decomposition, you can just write down the answer. The cover-up method is that simple. The differentiation becomes very easy because all the denominators are linear expressions with x-coefficients of one.

    BTW, it's always a good idea to check your partial fraction decomposition. One way (tedious) is to multiply everything out and verify that the expressions are algebraically the same. The other quick and dirty way is to use a transcendental value for x (like pi), plug it into the calculator. If the LHS comes out the same as the RHS (within the limits of round-off error), the decomposition *has* to be right.
    Last edited by a moderator: May 2, 2017
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