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[SOLVED] Any reason why this kind of problem is so hard?

DeusAbscondus

Active member
Jun 30, 2012
176
$$\frac{d}{dx} \frac{x}{\sqrt{x^2+3}}$$
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
 

DeusAbscondus

Active member
Jun 30, 2012
176
$$\frac{d}{dx} \frac{x}{\sqrt{x^2+3}}$$
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:

MarkFL

Pessimist Singularitarian
Staff member
Feb 24, 2012
13,646
St. Augustine, FL.
Applying the quotient rule, we find:

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

DeusAbscondus

Active member
Jun 30, 2012
176
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:

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Canada
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.
 

DeusAbscondus

Active member
Jun 30, 2012
176
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 .......(Nerd)
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Canada
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 .......(Nerd)
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.
 

DeusAbscondus

Active member
Jun 30, 2012
176
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)
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,176
Rochester, MN
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 come up from time to time.

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.
 

DeusAbscondus

Active member
Jun 30, 2012
176
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 come up from time to time.

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
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,176
Rochester, MN
Here's the correspondence of the mnemonic with the quotient rule ('dee' means 'take the derivative of'):

$$\text{dee}\,\left(\frac{\text{high}}{\text{low}}\right)=\frac{\text{low dee-high}-\text{high dee-low}}{\text{low}^{2}}.$$
"Low dee-high minus high dee-low over the square of what's below."

Just like

$$\frac{d}{dx}\left(\frac{h(x)}{\ell(x)}\right)= \frac{ \ell(x)\,\frac{d}{dx}\,h(x)-h(x)\,\frac{d}{dx}\,\ell(x)}{\ell^{2}(x)}.$$
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,176
Rochester, MN
Incidentally, you might also check into logarithmic differentiation for these kinds of problems. That might help a bit.
 

DeusAbscondus

Active member
Jun 30, 2012
176
Incidentally, you might also check into logarithmic differentiation for these kinds of problems. That might help a bit.
Thanks again.

Incidentally, I've just had considerably greater success at a heap of probs (admittedly, for the second or third time around for some of them), this time, though, using the quotient rule where appropriate.

Provisionally convinced. Suffice to say I henceforth carry it as one more tool in the kit!

DeusAbs