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Math Help - Differentiating a rational function

  1. #1
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    Differentiating a rational function

    Hello

    Excuse me if I make any improper use of mathematical experssions because English isn't my mother-tongue and I don't study math in English.

    I'm a high school student in the 10th. I have math exam tomorrow and while stuyding I came across this question where I must differentiate a rational function. The function is:

    y = 3 / ( (2x+1)^3 * sqrt(2x+1) )

    I solved it by first adding a minus, then squaring the denominator, then multiplying by the derivative of the denominator, which is 3.5*(2x+1)^2.5 * 2... This obviously gave me a correct answer...

    However, I recall that our teacher once mentioned a shortcut to derive a rational function, but gave us a simpler example. He said, in general:

    y = 1/x^n
    y' = - n / x^n+1

    But if I try to solve it that way, I get:

    y' = - 10.5 / (2x+1)^4.5, whilst the correct final derivative has 21 in the numerator. That means that I must multiply it by 2, which is the internal-derivative (dunno how u call it, but I'm referring to the 2x+1) of the denominator!

    So is that correct? Can I solve it in the above-mentioned method? If so, has the teacher actually missed that specific detail when he gave us the general law?

    Thanks in advance,
    Louie.
    Last edited by loui1410; June 4th 2007 at 02:34 AM.
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  2. #2
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    Quote Originally Posted by loui1410 View Post
    Hello

    ...
    The function is:

    y = 3 / ( (2x+1)^3 * sqrt(2x+1) )

    ...
    Hello,

    y = \frac{3}{(2x+1)^2 \cdot \sqrt{2x+1}} = 3 \cdot (2x+1)^{-\frac{7}{2}} Use chain rule to derivate:

    y' = 3 \cdot \left( -\frac{7}{2} \right) \cdot (2x+1)^{-\frac{7}{2} - 1} \cdot 2 which simplifies to:

    y' = -21 \cdot (2x+1)^{-\frac{9}{2}} = -\frac{21}{(2x+1)^4 \cdot \sqrt{2x+1}}
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  3. #3
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    Thanks but finding the derivative was not my question.
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  4. #4
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    Hello. This is probably a little late, but... You are asking whether you can apply the 'rule'

    y = \frac{1}{x^n} \Rightarrow y' \equiv \frac{dy}{dx} = \frac{-n}{x^{n+1}}

    to differentiate the function

    y = \frac{3}{(2x + 1)^3 \sqrt{2x + 1}} \equiv 3 (2x + 1)^{-7/2} ?

    The answer is no! Your function is not of the form 1 / x^n; it is of the form 1 / u^n where u is a function of x (here, u = (2x + 1)).

    In a problem like yours, where y is a function of u, and u is a function of x, you need to use a theorem called the chain rule. The chain rule states:

    y' \equiv \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} .

    So, you can use your 'rule' to find dy/du, but then you must multiply it by du/dx to find the true derivative.

    Just to give you the full calculation:

    <br />
y = \frac{3}{u^{7/2}} \Rightarrow \frac{dy}{du} = \frac{-3 \cdot (7/2)} {u^{9/2}} ~~; (simply using your rule!)

    <br />
u = 2x + 1 \Rightarrow \frac{du}{dx} = 2 ~~ ;<br />

    <br />
\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} = \frac{-3 \cdot (7/2)} {u^{9/2}} \cdot 2 = \frac{-21}{(2x+1)^{9/2}}.<br />

    I hope that helps.
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