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Derivative/chain rule

daigo

Member
Jun 27, 2012
60
[tex]f(x) = \frac{1}{x\sqrt{5-2x}}[/tex]
[tex]
f'(x) = (x\sqrt{5-2x})^{-1} \\
= -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot (x(5-2x)^{\frac{1}{2}})' \\
= -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((x') \cdot ((5-2x)^{\frac{1}{2}}) + ((x) \cdot ((5-2x)^{\frac{1}{2}})') \\
= -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((5-2x)^{\frac{1}{2}} + (x \cdot (\frac{1}{2}(5-2x)^{-\frac{1}{2}}) \cdot (5 - 2x)') \\
= -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((5-2x)^{\frac{1}{2}} + (x \cdot (\frac{1}{2}(5-2x)^{-\frac{1}{2}}) \cdot (- 2))) \\
= -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((5-2x)^{\frac{1}{2}} -x(5-2x)^{-\frac{1}{2}}) \\
= -\frac{(5-2x)^{\frac{1}{2}} -x(5-2x)^{-\frac{1}{2}}}{x^{2}(5-2x)} \\
[/tex]

Have I done anything wrong so far?
 

daigo

Member
Jun 27, 2012
60
After I simplify, I get this:

[tex]= \frac{(5 - 2x)^{\frac{1}{2}}}{x^{2}(5 - 2x)} - \frac{x(5 - 2x)^{-\frac{1}{2}}}{x^{2}(5 - 2x)} \\
\\
\\
= \frac{1}{x^{2}(5 - 2x)^{\frac{1}{2}}} - \frac{1}{x(5-2x)^{\frac{3}{2}}}[/tex]

But the solution says I'm wrong...what have I done?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
[tex]
f'(x) = (x\sqrt{5-2x})^{-1} \\
= -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot (x(5-2x)^{\frac{1}{2}})' \\
= -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((x') \cdot ((5-2x)^{\frac{1}{2}}) + ((x) \cdot ((5-2x)^{\frac{1}{2}})') \\
= -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((5-2x)^{\frac{1}{2}} + (x \cdot (\frac{1}{2}(5-2x)^{-\frac{1}{2}}) \cdot (5 - 2x)') \\
= -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((5-2x)^{\frac{1}{2}} + (x \cdot (\frac{1}{2}(5-2x)^{-\frac{1}{2}}) \cdot (- 2))) \\
= -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((5-2x)^{\frac{1}{2}} -x(5-2x)^{-\frac{1}{2}}) \\
= -\frac{(5-2x)^{\frac{1}{2}} -x(5-2x)^{-\frac{1}{2}}}{x^{2}(5-2x)} \\
[/tex]

Have I done anything wrong so far?
Hi diago, :)

Up to this point you are correct.

After I simplify, I get this:

[tex]= \frac{(5 - 2x)^{\frac{1}{2}}}{x^{2}(5 - 2x)} - \frac{x(5 - 2x)^{-\frac{1}{2}}}{x^{2}(5 - 2x)} \\
\\
\\
= \frac{1}{x^{2}(5 - 2x)^{\frac{1}{2}}} - \frac{1}{x(5-2x)^{\frac{3}{2}}}[/tex]

But the solution says I'm wrong...what have I done?
Note that, you have missed the minus sign before the fraction.

Kind Regards,
Sudharaka.
 

daigo

Member
Jun 27, 2012
60
I can't see where I've missed any negative sign
 

daigo

Member
Jun 27, 2012
60
Oh wait, I see it now...the very first negative sign at the very beginning...I completely missed that this entire time and I couldn't see it...wow
 

CaptainBlack

Well-known member
Jan 26, 2012
890
[tex]f(x) = \frac{1}{x\sqrt{5-2x}}[/tex]
You are making a meal of this. Rewrite as:

\[ f(x) = x^{-1}(5-2x)^{-1/2}\]
then using the product rule:
\[\begin{aligned}f'(x)&=(-1)x^{-2}(5-2x)^{-1/2}+x^{-1}(-1/2)(5-2x)^{-3/2}(-2)\\ \\ &=-x^{-1}f(x)+f(x)(5-2x)\\ \\&=f(x)\left[-\frac{1}{x}+\frac{1}{5-2x} \right]\\ \\ &=f(x)\left[ \frac{3x-5}{x(5-2x)}\right] \\ \\ &= \frac{3x-5} {x^2(5-2x)^{3/2}}\end{aligned}\]
 

DeusAbscondus

Active member
Jun 30, 2012
176
CB: strategy at line 3; please unpick this for me

You are making a meal of this. Rewrite as:

\[ f(x) = x^{-1}(5-2x)^{-1/2}\]
then using the product rule:
\[\begin{aligned}f'(x)&=(-1)x^{-2}(5-2x)^{-1/2}+x^{-1}(-1/2)(5-2x)^{-3/2}(-2)\\ \\ &=-x^{-1}f(x)+f(x)(5-2x)\\ \\&=f(x)\left[-\frac{1}{x}+\frac{1}{5-2x} \right]\\ \\ &=f(x)\left[ \frac{3x-5}{x(5-2x)}\right] \\ \\ &= \frac{3x-5} {x^2(5-2x)^{3/2}}\end{aligned}\]
Hello CB,
I note with interest the 2nd line of your calculations after: "using the product rule", which is
$$=x^{-1}(f(x)+f(x)(5-2x)$$
Would you kindly:
1) explain what is behind this move?
It looks like you are factoring out f(x) from both sides of the addition. Is that right? If so, could you enlighten me as to the arithmetic/logic involved?
(Looks like whatever it is, they didn't show us in class the other day when introducing the product rule, and looks like this would make these problems a hell of a lot easier to deal with)

(Edited 1 hour later) I have just surprised myself to see how easy, pleasant and fun it was to apply basic algebra and advance myself towards an answer to question (1) above:
(Edited 1 hour 30 mins from original post) {On further testing my hypothesis, set out below, I realize it is wrong, making what follows redundant, but perhaps not entirely: I will let it stand as an example of "going off half-cocked" with the result: shot in the pants!}
It goes like this (plse comment, emend, correct, contradict as and where appropriate (Nerd) )

Let $x^{-1}=u$
and $(5-2x)^{-1/2}=v$
Since $f'(x)=(u'v)+(uv')$ then, your equation $f'(x)=\left(x^-1(f(x))+(f(x)\times(5-2x)\right)$ implies that $f'(x)=u^2v+v^2u$ which can then be factored as:
$(uv)(u+v)$ (a very interesting result, if my reasoning is correct, which no-one told me about at school (Headbang))

Now,since $f(x)=(uv)$ this explains why you were able to write $f'(x)$ as the sum of $u(uv)$ and $v(vu)$ at line 2 of your calculations.

Am I right?
If so, it is then simply a matter of factoring $f(x)$ outside square brackets in the next line. And, if I am right, would this be the way to go as a point of "best practice" when working on all such problems?

Finally, if I have, if fact, anwered question (1) please go ahead and answer the next 2 typographical questions, or direct me to a convenient and usable (searchable) source of such short-cuts:

2) explain the short-cut I see in your latex "&="
3) tell me how to make a "hotkey" for things like "\begin{aligned}" so as to avoid having to type it out in full every time I want to use it


(Edited 1 hour 30 mins from original post) {On further testing my hypothesis, set out above, I realize it is wrong, making it largely redundant, but perhaps not entirely: I will let it stand as an example of "going off half-cocked" with the result: shot in the pants!}

Regs,
DeusAbs
 
Last edited:

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
[tex]
f'(x) = (x\sqrt{5-2x})^{-1} \\
= -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot (x(5-2x)^{\frac{1}{2}})' \\
= -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((x') \cdot ((5-2x)^{\frac{1}{2}}) + ((x) \cdot ((5-2x)^{\frac{1}{2}})') \\
= -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((5-2x)^{\frac{1}{2}} + (x \cdot (\frac{1}{2}(5-2x)^{-\frac{1}{2}}) \cdot (5 - 2x)') \\
= -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((5-2x)^{\frac{1}{2}} + (x \cdot (\frac{1}{2}(5-2x)^{-\frac{1}{2}}) \cdot (- 2))) \\
= -(x(5-2x)^{\frac{1}{2}})^{-2} \cdot ((5-2x)^{\frac{1}{2}} -x(5-2x)^{-\frac{1}{2}}) \\
= -\frac{(5-2x)^{\frac{1}{2}} -x(5-2x)^{-\frac{1}{2}}}{x^{2}(5-2x)} \\
[/tex]

Have I done anything wrong so far?
At the risk of adding too many cooks, I would say that the quotient rule is not a bad way to go here. The nice thing about the quotient rule is that it sometimes saves algebra after taking all your derivatives, because you don't have to get a common denominator.
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,042
DeusAbcondus,

I think it's really great you're diving into math this way of trying out new things and not memorizing "the book" to get by. You're already way ahead of most who encounter math in some way.

This line, \(\displaystyle f'(x)=u^2v+v^2u\) is not generally true for \(\displaystyle f(x)=u(x) \cdot v(x)\) (note that u and v are both themselves functions of x). You can in theory always factor out f(x) from something, but many times it won't do you any good. CB simply noticed that in this problem it would save time by doing so.

For example, if you have \(\displaystyle f(x)=x^3+2\) then \(\displaystyle f'(x)=3x^2\). You could write this as \(\displaystyle f(x) \cdot \frac{3x^2}{x^3+2}\) ,which is "factoring out" f(x) in a sense, but that would be silly. So factoring f(x) out for simplification purposes seems to be a situational trick to apply.

I hope this addresses what you were thinking of somewhat. Please let me know if I'm not getting your comments properly.
 

DeusAbscondus

Active member
Jun 30, 2012
176
DeusAbcondus,

I think it's really great you're diving into math this way of trying out new things and not memorizing "the book" to get by. You're already way ahead of most who encounter math in some way.

This line, \(\displaystyle f'(x)=u^2v+v^2u\) is not generally true for \(\displaystyle f(x)=u(x) \cdot v(x)\) (note that u and v are both themselves functions of x). You can in theory always factor out f(x) from something, but many times it won't do you any good. CB simply noticed that in this problem it would save time by doing so.

For example, if you have \(\displaystyle f(x)=x^3+2\) then \(\displaystyle f'(x)=3x^2\). You could write this as \(\displaystyle f(x) \cdot \frac{3x^2}{x^3+2}\) ,which is "factoring out" f(x) in a sense, but that would be silly. So factoring f(x) out for simplification purposes seems to be a situational trick to apply.

I hope this addresses what you were thinking of somewhat. Please let me know if I'm not getting your comments properly.
Thanks Jameson, not least for your encouraging (and very generous) appraisal of my faltering steps.
I was momentarily stunned and delighted by that move by CB, which I'd not encountered before (early days for me) so I leapt upon it like a treasure-seeker, thinking it was going to prove to be a key to a treasure-trove!
However, that still leaves me in a quandary as to just what he did do by that step: I still can't see it, as if it is a conjurer's trick with the pea under one of three cups, or none at all!

Having been so good to read this far, could you put me out of my misery and explain a bit further that particular line of his reasoning?

Thanks ever so much,
D'abs
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,042
I'll do my best :) In all honesty I'm not really one of the "math guys" here. I handle all of the administrative ins and outs to keep the site running and promote it. Of course I have a certain passion for math, otherwise I wouldn't be here, but my level isn't as high as many others.

Ok, so let's start at the beginning.

CB rewrote \(\displaystyle f(x) = \frac{1}{x\sqrt{5-2x}} \text{ as } f(x) = x^{-1}(5-2x)^{-1/2}\). I think that should be clear.

Then he took the derivative using the product rule. If y = uv then y' = uv'+u'v. It looks like the way he wrote it though he just reversed the order and did u'v+uv' but that's the same result naturally. That's this line.

\(\displaystyle f'(x)=(-1)x^{-2}(5-2x)^{-1/2}+x^{-1}(-1/2)(5-2x)^{-3/2}(-2)\)

Now comes the factoring part. There are two main terms in the derivative separated by an addition sign. Let's look at the first.

\(\displaystyle (-1)x^{-2}(5-2x)^{-1/2}\) <- this term. Remember that \(\displaystyle f(x)=x^{-1}(5-2x)^{-1/2}\)

When looking at the term you'll notice many similarities. The \(\displaystyle (5-2x)^{-1/2}\) is exactly the same and the \(\displaystyle -x^{-2}\) is not that different from \(\displaystyle x^{-1}\) which is in f(x). Here is where the factoring begins.

This uses a very common and powerful rule of combining exponents which are multiplied together that says \(\displaystyle x^a \cdot x^b=x^{a+b}\) Using that in our case we note that \(\displaystyle x^{-2}=x^{-1} \cdot x^{-1}\).

So finally we start with \(\displaystyle (-1)x^{-2}(5-2x)^{-1/2}=(-1) \cdot \left( x^{-1} \cdot x^{-1} \right) \cdot (5-2x)^{-1/2}\)

Now we just regroup the parentheses, getting \(\displaystyle \left( (-1) \cdot x^{-1} \right) \cdot \left( x^{-1} \cdot (5-2x)^{-1/2} \right)= \left( (-1) \cdot x^{-1} \right) \cdot f(x)=-x^{-1} \cdot f(x)\)

Is this along the lines of what you were asking?
 

DeusAbscondus

Active member
Jun 30, 2012
176
I'll do my best :)
This uses a very common and powerful rule of combining exponents which are multiplied together that says \(\displaystyle x^a \cdot x^b=x^{a+b}\) Using that in our case we note that \(\displaystyle x^{-2}=x^{-1} \cdot x^{-1}\).

So finally we start with \(\displaystyle (-1)x^{-2}(5-2x)^{-1/2}=(-1) \cdot \left( x^{-1} \cdot x^{-1} \right) \cdot (5-2x)^{-1/2}\)

Now we just regroup the parentheses, getting \(\displaystyle \left( (-1) \cdot x^{-1} \right) \cdot \left( x^{-1} \cdot (5-2x)^{-1/2} \right)= \left( (-1) \cdot x^{-1} \right) \cdot f(x)=-x^{-1} \cdot f(x)\)

Is this along the lines of what you were asking?
Jameson,
Not only does it hit the nail squarely and sweetly on the spot, it does so with concision and elegance.
I love maths and I love this site that puts me in touch with excellent minds and good-hearted people, and sometimes, with these qualities mixed in the same person, such as in your case.

Big thanks and big respect!
DeusAbs
 

CaptainBlack

Well-known member
Jan 26, 2012
890
At the risk of adding too many cooks, I would say that the quotient rule is not a bad way to go here. The nice thing about the quotient rule is that it sometimes saves algebra after taking all your derivatives, because you don't have to get a common denominator.
There is an excellent reason why I did not even consider using the quotient rule; that is that I do not know it. Why do I not know it you ask, well when we did it when I was 16 or so I realised that it was redundant, every thing that can be done with the quotient rule can be done with the product rule since:

\[\frac{d}{dx}\left( \frac{u(x)}{v(x)}\right)=\frac{d}{dx}\left( u(x) \times (v(x))^{-1}\right)=u'v^{-1}-u v^{-2}v'\]

by the product rule, which saves valuable wet storage.

CB