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    #1
    $$\frac{d}{dx} \frac{x}{\sqrt{x^2+3}}$$
    I've been trying to get this for an hour now and I just get a mess; could anyone lend me a hand to see where I'm going wrong?

    thx,
    DeusAbs

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    #2 Thread Author
    Quote Originally Posted by DeusAbscondus View Post
    $$\frac{d}{dx} \frac{x}{\sqrt{x^2+3}}$$
    I've been trying to get this for an hour now and I just get a mess; could anyone lend me a hand to see where I'm going wrong?

    thx,
    DeusAbs
    I've tried it this way:
    $$y'=x(x^2+3)^{-1/2}$$
    $$=1(x^2+3)^{-1/2}+x\cdot -\frac{1}{2}(x^2+3)^{-3/2}\cdot 2x$$
    $$=(x^2+3)^{-1/2}+x\cdot -\frac{2x}{2}(x^2+3)^{-3/2}$$
    $$=(x^2+3)^{-1/2}+x^2(x^2+3)^{-3/2}$$

    starting to look messy and i've got a feeling that I've made the same mistake somewhere but can;'t see it
    Last edited by DeusAbscondus; October 9th, 2012 at 06:02.

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    #3
    Applying the quotient rule, we find:

    $\displaystyle \frac{d}{dx}\left(\frac{x}{\sqrt{x^2+3}} \right)=\frac{\sqrt{x^2+3}(1)-x\left(\frac{x}{\sqrt{x^2+3}} \right)}{x^2+3}=\frac{x^2+3-x^2}{(x^2+3)^{\frac{3}{2}}}=\frac{3}{(x^2+3)^{ \frac{3}{2}}}$

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    #4 Thread Author
    Quote Originally Posted by MarkFL View Post
    Applying the quotient rule, we find:

    $\displaystyle \frac{d}{dx}\left(\frac{x}{\sqrt{x^2+3}} \right)=\frac{\sqrt{x^2+3}(1)-x\left(\frac{x}{\sqrt{x^2+3}} \right)}{x^2+3}=\frac{x^2+3-x^2}{(x^2+3)^{\frac{3}{2}}}=\frac{3}{(x^2+3)^{ \frac{3}{2}}}$

    right, but is there any reasonable in principle why the product rule shouldn't work pretty straightforwardly in this case?

    i'm going to continue with product rule, as I havent' come across an argument to best CB's which goes like this: "why carry around in one's head a single redundant formula? The quotient rule seems redundant to me, thus I proceed doggedly:

    $$y'=x(x^2+3)^{-1/2}$$
    $$=1(x^2+3)^{-1/2}+x\cdot -\frac{1}{2}(x^2+3)^{-3/2}\cdot 2x$$
    $$=(x^2+3)^{-1/2}+x\cdot -\frac{2x}{2}(x^2+3)^{-3/2}$$
    $$=(x^2+3)^{-1/2}+x^2(x^2+3)^{-3/2}$$
    $$=\frac{1}{(x^2+3)^{1/2}}-\frac{x^2}{(x^2+3)^{3/2}}=\frac{(x^2+3-x^2)}{(x^2+3)^{3/2}}=\frac{3}{(x^2+3)^{3/2}}$$

    Voila! Je l'ai eu, ce fils de satan!
    Last edited by DeusAbscondus; October 9th, 2012 at 07:48.

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    #5
    Quote Originally Posted by DeusAbscondus View Post
    right, but is there any reasonable in principle why the product rule shouldn't work pretty straightforwardly in this case?

    i'm going to continue with product rule, as I havent' come across an argument to best CB's which goes like this: "why carry around in one's head a single redundant formula? The quotient rule seems redundant to me, thus I proceed doggedly:
    Hi DeusAbscondus,

    The way you have tried to do the problem is correct but you can simplify further.

    \begin{eqnarray}

    \frac{d}{dx}\left(\frac{x}{\sqrt{x^2+3}} \right)&=&\frac{d}{dx}\left[x(x^2+3)^{-1/2}\right]\\

    &=&(x^2+3)^{-1/2}+x\left[-\frac{1}{2}.(x^2+3)^{-3/2}.2x\right]\\

    &=&\frac{1}{(x^2+3)^{1/2}}-\frac{x^2}{(x^2+3)^{3/2}}\\

    &=&\frac{(x^2+3)-x^2}{(x^2+3)^{3/2}}\\

    &=&\frac{3}{(x^2+3)^{3/2}}

    \end{eqnarray}

    Kind Regards,
    Sudharaka.

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    #6 Thread Author
    Quote Originally Posted by Sudharaka View Post
    Hi DeusAbscondus,

    The way you have tried to do the problem is correct but you can simplify further.

    \begin{eqnarray}

    \frac{d}{dx}\left(\frac{x}{\sqrt{x^2+3}} \right)&=&\frac{d}{dx}\left[x(x^2+3)^{-1/2}\right]\\

    &=&(x^2+3)^{-1/2}+x\left[-\frac{1}{2}.(x^2+3)^{-3/2}.2x\right]\\

    &=&\frac{1}{(x^2+3)^{1/2}}-\frac{x^2}{(x^2+3)^{3/2}}\\

    &=&\frac{(x^2+3)-x^2}{(x^2+3)^{3/2}}\\

    &=&\frac{3}{(x^2+3)^{3/2}}

    \end{eqnarray}

    Kind Regards,
    Sudharaka.
    Thanks kindly Sudharaka; I got there just as I saw your post come up!
    I don't know why I have such persistent trouble with these; but there is no way back now! I will just have to get there by dogged persistence, as I am too close to becoming a Time Lord when I come to master the Integral .......

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    #7
    Quote Originally Posted by DeusAbscondus View Post
    Thanks kindly Sudharaka; I got there just as I saw your post come up!
    I don't know why I have such persistent trouble with these; but there is no way back now! I will just have to get there by dogged persistence, as I am too close to becoming a Time Lord when I come to master the Integral .......
    You are welcome. Note that the Quotient rule can be derived from the product rule. In fact whenever you have a quotient of two functions \(\frac{f}{g}\) you can view it as a product of \(f\) and \(\frac{1}{g}\). So using the product rule you can get,

    \begin{eqnarray}

    \left(\frac{f}{g}\right)'&=&\frac{f'}{g}+f\left( \frac{1}{g}\right)'\\

    &=&\frac{f'}{g}-\frac{fg'}{g^2}\\

    &=&\frac{f'g-fg'}{g^2}\\

    \end{eqnarray}

    which is the quotient rule.

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    #8 Thread Author
    Quote Originally Posted by Sudharaka View Post
    You are welcome. Note that the Quotient rule can be derived from the product rule. In fact whenever you have a quotient of two functions \(\frac{f}{g}\) you can view it as a product of \(f\) and \(\frac{1}{g}\). So using the product rule you can get,

    \begin{eqnarray}

    \left(\frac{f}{g}\right)'&=&\frac{f'}{g}+f\left( \frac{1}{g}\right)'\\

    &=&\frac{f'}{g}-\frac{fg'}{g^2}\\

    &=&\frac{f'g-fg'}{g^2}\\

    \end{eqnarray}

    which is the quotient rule.
    Thanks Sudharaka. Going to write this up immediately in my memory cards.
    (Using geogebra's latex facility to do this conveniently and neatly, with graphs sketched in to go with the calculations, after much "shopping around" and trying different software)

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    #9
    Quote Originally Posted by DeusAbscondus View Post
    right, but is there any reasonable in principle why the product rule shouldn't work pretty straightforwardly in this case?

    i'm going to continue with product rule, as I havent' come across an argument to best CB's which goes like this: "why carry around in one's head a single redundant formula? The quotient rule seems redundant to me...
    Ah, but it's not, at least not to my mind. The quotient rule allows you to avoid finding the common denominator (introducing possibilities for error), as well as giving you an integration formula that, while not used often, has .

    So the product-rule-only approach does allow you to remember one less formula, at the cost of more algebra. The quotient-rule approach allows you to do less algebra, at the cost of remembering one more formula. But the formula has a nice mnemonic device: "low dee-high minus high dee-low over the square of what's below." So it's not too bad.

    I'm in favor of the quotient rule, because I'd rather do more calculus and less algebra.

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    #10 Thread Author
    Quote Originally Posted by Ackbach View Post
    Ah, but it's not, at least not to my mind. The quotient rule allows you to avoid finding the common denominator (introducing possibilities for error), as well as giving you an integration formula that, while not used often, has .

    So the product-rule-only approach does allow you to remember one less formula, at the cost of more algebra. The quotient-rule approach allows you to do less algebra, at the cost of remembering one more formula. But the formula has a nice mnemonic device: "low dee-high minus high dee-low over the square of what's below." So it's not too bad.

    I'm in favor of the quotient rule, because I'd rather do more calculus and less algebra.

    Ackbach, I love this. Thanks.
    If you get a chance, please render your mnemonic more explicit, and I'll practice it, use it and see if I can see what you are getting at.
    I have a heap of such problems I'm working through now, ALL by product rule, and, it must be admitted, with mistakes a plenty. But I have never worked with quotient rule and though I know it, I just can't see how the mnemonic captures it.

    Greatly appreciate your responses,
    DeusAbs

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