MHB Understanding the Chain Rule and Product Rule in Multivariable Calculus

[Petrus]

Hello MHB,
I got one exempel that I don't get same result as my book.
Exempel: If $$z=f(x,y)$$ has continuos second-order partial derivates and $$x=r^2+s^2$$ and $$y=2rs$$ find $$\frac{d^2z}{dr^2}$$
So what I did before checking soulotion:
$$\frac{d^2z}{dr^2}=\frac{dz}{dr} \frac{d}{dr}$$
So I start with solving $$\frac{dz}{dr}=\frac{dz}{dx}\frac{dx}{dr}+\frac{dz}{dy}\frac{dy}{dr} = \frac{dz}{dx}(2r)+\frac{dz}{dy}(2s)$$
so now we got $$\frac{d}{dr}(\frac{dz}{dx}(2r)+\frac{dz}{dy}(2s))$$ and that is equal to $$2r\frac{d}{dr}\frac{dz}{dx}+2s\frac{d}{dr}\frac{dz}{dy}$$
my book soloution:
$$\frac{d^2z}{dr^2}= \frac{d}{dr}(\frac{dz}{dx}(2r)+\frac{dz}{dy}(2s))$$ and they equal that to $$2\frac{dz}{dx}+2r\frac{d}{dr}(\frac{dz}{dx})+2s \frac{d}{dr}(\frac{dz}{dy})$$
my question is where do they get that extra $$2\frac{dz}{dx}$$ (I suspect it was a accident) and my last question why do they got parantes on $$(\frac{dz}{dx})$$ and $$(\frac{dz}{dy})$$
then they solve $$\frac{d}{dr} (\frac{dz}{dx})$$ and $$\frac{d}{dr}(\frac{dz}{dx})$$
why do they do that and how do they do it? unfortently I have to run so I can't post fully soloution, will post it later.

Regards,

 

[alyafey22]

Re: chain rule and product rule

so now we got $$\frac{d}{dr}(\frac{dz}{dx}(2r)+\frac{dz}{dy}(2s))$$ and that is equal to $$2r\frac{d}{dr}\frac{dz}{dx}+2s\frac{d}{dr}\frac{dz}{dy}$$

here you have a mistake $$2r $$ can't be taken out , it is not constant .

Spoiler

use product rule

 

[Petrus]

Re: chain rule and product rule

here you have a mistake $$2r $$ can't be taken out , it is not constant .

ignore thanks zaid

 

[Petrus]

Re: chain rule and product rule

here you have a mistake $$2r $$ can't be taken out , it is not constant .

Spoiler

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?

 

[alyafey22]

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]

I don't understand this part
$$\frac{d}{dr}(\frac{dz}{dx})=\frac{d}{dx}(\frac{dz}{dx})\frac{dx}{dr}+(\frac{d}{dy}(\frac{dz}{dx}) \frac{dy}{dr}=\frac{d^z}{dz^2}(2r)+\frac{d^2z}{dydx}(2s)$$
they do same with $$\frac{d}{dr}( \frac{dz}{dy})$$ and I don't understand it, what does it means to take $$\frac{d}{dr}(\frac{dz}{dx})$$

Regards,

 

[alyafey22]

We know that :

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

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

so what do we do using the chain rule ?

 

[alyafey22]

Here is a hint

Spoiler

substitute $$\frac{dz}{dx}$$ instead of each $$z$$

 

[Petrus]

Here is a hint

Spoiler

substitute $$\frac{dz}{dx}$$ instead of each $$z$$

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!
$$\frac{d}{dr}(z) = \frac{dz}{dx}\cdot \frac{dx}{dr}+\frac{dz}{dy}\cdot \frac{dy}{dr} <=> \frac{d}{dr}\frac{dz}{dx}=\frac{d(\frac{dz}{dx})}{dx}\frac{dx}{dr}+ \frac{d( \frac{dz}{dx})}{dy}\frac{dy}{dr} <=> \frac{d}{dr}\frac{dz}{dx}=\frac{\frac{d^2z}{dx}}{dx}\frac{dx}{dr}+ \frac{ \frac{d^2z}{dx}}{dy}\frac{dy}{dr} $$ and now we can use the rule $$\frac{ \frac{a}{b}}{\frac{c}{d}} = \frac{ad}{bc}$$ to make it easy to see I always add an /1 in evry term in bottom exempel $$dx= \frac{dx}{1}$$(I somehow screw it when I try it with latex :S) so we got $$\frac{d}{dr}\frac{dz}{dx}=\frac{d^2z}{(dx)^2}\frac{dx}{dr}+ \frac{d^2z}{dxdy}\frac{dy}{dr} <=>\frac{d}{dr}\frac{dz}{dx}= \frac{d}{dx}\frac{dz}{dx}\frac{dx}{dr}+\frac{d}{dx}\frac{dz}{dy}\frac{dy}{dr}$$

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

 

[alyafey22]

$$\frac{d}{dr}(z) = \frac{dz}{dx}\cdot \frac{dx}{dr}+\frac{dz}{dy}\cdot \frac{dy}{dr} <=> \frac{d}{dr}\frac{dz}{dx}=\frac{d(\frac{dz}{dz})}{dx}\frac{dx}{dr}+ \frac{d( \frac{dz}{dx})}{dy}\frac{dy}{dr} $$

You have some mistakes , how did you get $$\frac{dz}{dz}$$?

 

[Petrus]

You have some mistakes , how did you get $$\frac{dz}{dz}$$?

Typo... I am working with edit it... got confused:S

 

[Petrus]

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)

 

[alyafey22]

You have some wrong notations

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

what is that equal to ?

 

[Petrus]

You have some wrong notations

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

what is that equal to ?

$$\frac{d^2z}{(dx)^2}$$

 

[alyafey22]

$$\frac{d^2z}{(dx)^2}$$

This is the second derivative of z with respect to x , it isn't usual multiplication !

 

[Petrus]

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... $$\frac{d^2z}{dx^2}$$

 

[alyafey22]

I always have write it as that :s but I think in my brain that it means second derivate but that is obvious wrong... $$\frac{d^2z}{dx^2}$$

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 .