
MHB Craftsman
#1
October 9th, 2012,
05:41
$$\frac{d}{dx} \frac{x}{\sqrt{x^2+3}}$$
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

October 9th, 2012 05:41
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MHB Craftsman
#2
October 9th, 2012,
05:46
Thread Author
Originally Posted by
DeusAbscondus
$$\frac{d}{dx} \frac{x}{\sqrt{x^2+3}}$$
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.

Pessimist Singularitarian
#3
October 9th, 2012,
05:53
Applying the quotient rule, we find:
$\displaystyle \frac{d}{dx}\left(\frac{x}{\sqrt{x^2+3}} \right)=\frac{\sqrt{x^2+3}(1)x\left(\frac{x}{\sqrt{x^2+3}} \right)}{x^2+3}=\frac{x^2+3x^2}{(x^2+3)^{\frac{3}{2}}}=\frac{3}{(x^2+3)^{ \frac{3}{2}}}$

MHB Craftsman
#4
October 9th, 2012,
06:16
Thread Author
Originally Posted by
MarkFL
Applying the quotient rule, we find:
$\displaystyle \frac{d}{dx}\left(\frac{x}{\sqrt{x^2+3}} \right)=\frac{\sqrt{x^2+3}(1)x\left(\frac{x}{\sqrt{x^2+3}} \right)}{x^2+3}=\frac{x^2+3x^2}{(x^2+3)^{\frac{3}{2}}}=\frac{3}{(x^2+3)^{ \frac{3}{2}}}$
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:
$$y'=x(x^2+3)^{1/2}$$
$$=1(x^2+3)^{1/2}+x\cdot \frac{1}{2}(x^2+3)^{3/2}\cdot 2x$$
$$=(x^2+3)^{1/2}+x\cdot \frac{2x}{2}(x^2+3)^{3/2}$$
$$=(x^2+3)^{1/2}+x^2(x^2+3)^{3/2}$$
$$=\frac{1}{(x^2+3)^{1/2}}\frac{x^2}{(x^2+3)^{3/2}}=\frac{(x^2+3x^2)}{(x^2+3)^{3/2}}=\frac{3}{(x^2+3)^{3/2}}$$
Voila! Je l'ai eu, ce fils de satan!
Last edited by DeusAbscondus; October 9th, 2012 at 07:48.

MHB Master
#5
October 9th, 2012,
07:34
Originally Posted by
DeusAbscondus
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.

MHB Craftsman
#6
October 9th, 2012,
07:51
Thread Author
Originally Posted by
Sudharaka
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 .......

MHB Master
#7
October 9th, 2012,
08:14
Originally Posted by
DeusAbscondus
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'gfg'}{g^2}\\
\end{eqnarray}
which is the quotient rule.

MHB Craftsman
#8
October 9th, 2012,
08:25
Thread Author
Originally Posted by
Sudharaka
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'gfg'}{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)

#9
October 9th, 2012,
13:37
Originally Posted by
DeusAbscondus
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 productruleonly approach does allow you to remember one less formula, at the cost of more algebra. The quotientrule approach allows you to do less algebra, at the cost of remembering one more formula. But the formula has a nice mnemonic device: "low deehigh minus high deelow 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.

MHB Craftsman
#10
October 10th, 2012,
02:31
Thread Author
Originally Posted by
Ackbach
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 productruleonly approach does allow you to remember one less formula, at the cost of more algebra. The quotientrule approach allows you to do less algebra, at the cost of remembering one more formula. But the formula has a nice mnemonic device: "low deehigh minus high deelow 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