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Chain rule and product rule

Petrus

Well-known member
Feb 21, 2013
739
Hello MHB,
I got one exempel that I dont get same result as my book.
Exempel: If \(\displaystyle z=f(x,y)\) has continuos second-order partial derivates and \(\displaystyle x=r^2+s^2\) and \(\displaystyle y=2rs\) find \(\displaystyle \frac{d^2z}{dr^2}\)
So what I did before checking soulotion:
\(\displaystyle \frac{d^2z}{dr^2}=\frac{dz}{dr} \frac{d}{dr}\)
So I start with solving \(\displaystyle \frac{dz}{dr}=\frac{dz}{dx}\frac{dx}{dr}+\frac{dz}{dy}\frac{dy}{dr} = \frac{dz}{dx}(2r)+\frac{dz}{dy}(2s)\)
so now we got \(\displaystyle \frac{d}{dr}(\frac{dz}{dx}(2r)+\frac{dz}{dy}(2s))\) and that is equal to \(\displaystyle 2r\frac{d}{dr}\frac{dz}{dx}+2s\frac{d}{dr}\frac{dz}{dy}\)
my book soloution:
\(\displaystyle \frac{d^2z}{dr^2}= \frac{d}{dr}(\frac{dz}{dx}(2r)+\frac{dz}{dy}(2s))\) and they equal that to \(\displaystyle 2\frac{dz}{dx}+2r\frac{d}{dr}(\frac{dz}{dx})+2s \frac{d}{dr}(\frac{dz}{dy})\)
my question is where do they get that extra \(\displaystyle 2\frac{dz}{dx}\) (I suspect it was a accident) and my last question why do they got parantes on \(\displaystyle (\frac{dz}{dx})\) and \(\displaystyle (\frac{dz}{dy})\)
then they solve \(\displaystyle \frac{d}{dr} (\frac{dz}{dx})\) and \(\displaystyle \frac{d}{dr}(\frac{dz}{dx})\)
why do they do that and how do they do it? unfortently I have to run so I cant post fully soloution, will post it later.

Regards,
 
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ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: chain rule and product rule

so now we got \(\displaystyle \frac{d}{dr}(\frac{dz}{dx}(2r)+\frac{dz}{dy}(2s))\) and that is equal to \(\displaystyle 2r\frac{d}{dr}\frac{dz}{dx}+2s\frac{d}{dr}\frac{dz}{dy}\)
here you have a mistake \(\displaystyle 2r \) can't be taken out , it is not constant .

use product rule
 
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Petrus

Well-known member
Feb 21, 2013
739
Re: chain rule and product rule

here you have a mistake \(\displaystyle 2r \) can't be taken out , it is not constant .
ignore thanks zaid
 

Petrus

Well-known member
Feb 21, 2013
739
Re: chain rule and product rule

here you have a mistake \(\displaystyle 2r \) can't be taken out , it is not constant .

use product rule
I got one question. How do we know for sure that our function z got variable r in it? May sound stupid but I would like to know? Is it because we can rewrite x and y?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: chain rule and product rule

I got one question. How do we know for sure that our function z got variable r in it? May sound stupid but I would like to know? Is it because we can rewrite x and y?
We can't make sure , but it depends on r because both x and y depends on the value of r and s.
 

Petrus

Well-known member
Feb 21, 2013
739
I dont understand this part
\(\displaystyle \frac{d}{dr}(\frac{dz}{dx})=\frac{d}{dx}(\frac{dz}{dx})\frac{dx}{dr}+(\frac{d}{dy}(\frac{dz}{dx}) \frac{dy}{dr}=\frac{d^z}{dz^2}(2r)+\frac{d^2z}{dydx}(2s)\)
they do same with \(\displaystyle \frac{d}{dr}( \frac{dz}{dy})\) and I dont understand it, what does it means to take \(\displaystyle \frac{d}{dr}(\frac{dz}{dx})\)

Regards,
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
We know that :

\(\displaystyle \frac{d}{dr}(z) = \frac{dz}{dx}\cdot \frac{dx}{dr}+\frac{dz}{dy}\cdot \frac{dy}{dr}\)

but we want to find here \(\displaystyle \frac{d}{dr} \left( \frac{dz}{dx}\right)\)

so what do we do using the chain rule ?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Here is a hint

substitute \(\displaystyle \frac{dz}{dx}\) instead of each \(\displaystyle z\)
 

Petrus

Well-known member
Feb 21, 2013
739
Here is a hint

substitute \(\displaystyle \frac{dz}{dx}\) instead of each \(\displaystyle z\)
that made it more clear, I was like thinking and thinking and then when I read your hint it made clear!! I was never thinking about that!
\(\displaystyle \frac{d}{dr}(z) = \frac{dz}{dx}\cdot \frac{dx}{dr}+\frac{dz}{dy}\cdot \frac{dy}{dr} <=> \frac{d}{dr}\frac{dz}{dx}=\frac{d(\frac{dz}{dx})}{dx}\frac{dx}{dr}+ \frac{d( \frac{dz}{dx})}{dy}\frac{dy}{dr} <=> \frac{d}{dr}\frac{dz}{dx}=\frac{\frac{d^2z}{dx}}{dx}\frac{dx}{dr}+ \frac{ \frac{d^2z}{dx}}{dy}\frac{dy}{dr} \) and now we can use the rule \(\displaystyle \frac{ \frac{a}{b}}{\frac{c}{d}} = \frac{ad}{bc}\) to make it easy to see I always add an /1 in evry term in bottom exempel \(\displaystyle dx= \frac{dx}{1}\)(I somehow screw it when I try it with latex :S) so we got \(\displaystyle \frac{d}{dr}\frac{dz}{dx}=\frac{d^2z}{(dx)^2}\frac{dx}{dr}+ \frac{d^2z}{dxdy}\frac{dy}{dr} <=>\frac{d}{dr}\frac{dz}{dx}= \frac{d}{dx}\frac{dz}{dx}\frac{dx}{dr}+\frac{d}{dx}\frac{dz}{dy}\frac{dy}{dr}\)

(Huh was a challange to do this on \(\displaystyle \LaTeX\) but worked well:D)
 
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ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
\(\displaystyle \frac{d}{dr}(z) = \frac{dz}{dx}\cdot \frac{dx}{dr}+\frac{dz}{dy}\cdot \frac{dy}{dr} <=> \frac{d}{dr}\frac{dz}{dx}=\frac{d(\frac{dz}{dz})}{dx}\frac{dx}{dr}+ \frac{d( \frac{dz}{dx})}{dy}\frac{dy}{dr} \)
You have some mistakes , how did you get \(\displaystyle \frac{dz}{dz}\)?
 

Petrus

Well-known member
Feb 21, 2013
739

Petrus

Well-known member
Feb 21, 2013
739
Thanks Zaid:) I finish edit it and now I understand what my book did :) Unfortently I could not figoure it out by myself but now I learned it! Thanks for taking your time(Smile)(Yes)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
You have some wrong notations

\(\displaystyle \frac{d}{dx} \left( \frac{dz}{dx}\right)\)

what is that equal to ?
 

Petrus

Well-known member
Feb 21, 2013
739
You have some wrong notations

\(\displaystyle \frac{d}{dx} \left( \frac{dz}{dx}\right)\)

what is that equal to ?
\(\displaystyle \frac{d^2z}{(dx)^2}\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
\(\displaystyle \frac{d^2z}{(dx)^2}\)
This is the second derivative of z with respect to x , it isn't usual multiplication !
 

Petrus

Well-known member
Feb 21, 2013
739
This is the second derivative of z with respect to x , it isn't usual multiplication !
I always have write it as that :s but I think in my brain that it means second derivate but that is obvious wrong.... \(\displaystyle \frac{d^2z}{dx^2}\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
I always have write it as that :s but I think in my brain that it means second derivate but that is obvious wrong.... \(\displaystyle \frac{d^2z}{dx^2}\)
The most important thing that you understand , it is not a matter of notations it is a matter of comprehending the concept behind , best luck .