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The chain rule (General version)

Petrus

Well-known member
Feb 21, 2013
739
Exempel 6: If \(\displaystyle g(s, t) = f(s^2-t^2, t^2-s^2)\) and f is differentiable, show that g satisfies the equation
\(\displaystyle t\frac{dg}{ds}+s\frac{dg}{dt}=0\)
I always try solve it before I look 'soloution' so this is how we both did it. (remember this is a exempel in my book so they show how to solve it but they did not make any tree driagram etc.)
\(\displaystyle \frac{dg}{ds}=\frac{df}{dx}\frac{dx}{ds}+\frac{df}{dy}\frac{dy}{ds}=\frac{df}{dx}(2s)+\frac{df}{dy}(-2s)\)
\(\displaystyle \frac{dg}{dt}=\frac{df}{dx}\frac{dx}{dt}+\frac{df}{dy}\frac{dy}{dt}=\frac{df}{dx}(-2t)+\frac{df}{dy}(2t)\)
\(\displaystyle t\frac{dg}{ds}+s\frac{dg}{dt}=\frac{df}{dx}(2st)+ \frac{df}{dy}(-2st) - \frac{df}{dx}(2st)+\frac{df}{dy}(2st)=0\)
This is what I made extra cause I wanna be specefic. I wanna citat from my book:
"We got n variables (\(\displaystyle x_1,x_2,....x_n\) (in this problem we got 2) and each of these \(\displaystyle x_j\) is a differentiable function of the m vavariables \(\displaystyle t_1, t_2,.....,t_m\) then u is a function of \(\displaystyle t_1,t_2,....,t_m\) (in our it will be 2 if I understand correct but I strugle on that one)"
So now I did write about these note and got same answer as the soloution. Then I wanted to draw a 'tree driagram' and I did draw it like this, correct me if I do it wrong math1.png

Regards,
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Exempel 6: If \(\displaystyle g(s, t) = f(s^2-t^2, t^2-s^2)\) and f is differentiable, show that g satisfies the equation
\(\displaystyle t\frac{dg}{ds}+s\frac{dg}{dt}=0\)
I always try solve it before I look 'soloution' so this is how we both did it. (remember this is a exempel in my book so they show how to solve it but they did not make any tree driagram etc.)
\(\displaystyle \frac{dg}{ds}=\frac{df}{dx}\frac{dx}{ds}+\frac{df}{dy}\frac{dy}{ds}=\frac{df}{dx}(2s)+\frac{df}{dy}(-2s)\)
\(\displaystyle \frac{dg}{dt}=\frac{df}{dx}\frac{dx}{dt}+\frac{df}{dy}\frac{dy}{dt}=\frac{df}{dx}(-2t)+\frac{df}{dy}(2t)\)
\(\displaystyle t\frac{dg}{ds}+s\frac{dg}{dt}=\frac{df}{dx}(2st)+ \frac{df}{dy}(-2st) - \frac{df}{dx}(2st)+\frac{df}{dy}(2st)=0\)
This is what I made extra cause I wanna be specefic. I wanna citat from my book:
"We got n variables (\(\displaystyle x_1,x_2,....x_n\) (in this problem we got 2) and each of these \(\displaystyle x_j\) is a differentiable function of the m vavariables \(\displaystyle t_1, t_2,.....,t_m\) then u is a function of \(\displaystyle t_1,t_2,....,t_m\) (in our it will be 2 if I understand correct but I strugle on that one)"
So now I did write about these note and got same answer as the soloution. Then I wanted to draw a 'tree driagram' and I did draw it like this, correct me if I do it wrong View attachment 735

Regards,
Hi Petrus, :)

Not sure whether you still need to clarify this question but here goes. Everything is correct, except the chain rule for multivariable functions involve partial derivatives; so I would have written,

\[\frac{\partial g}{\partial s}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial s}=2s\frac{\partial f}{dx}-2s\frac{\partial f}{\partial y}\]

and so on. The tree diagram is also correct. A nice reference about the chain rule with similar kind of examples is,

Pauls Online Notes : Calculus III - Chain Rule
 

Petrus

Well-known member
Feb 21, 2013
739
Hi Petrus, :)

Not sure whether you still need to clarify this question but here goes. Everything is correct, except the chain rule for multivariable functions involve partial derivatives; so I would have written,

\[\frac{\partial g}{\partial s}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial s}=2s\frac{\partial f}{dx}-2s\frac{\partial f}{\partial y}\]

and so on. The tree diagram is also correct. A nice reference about the chain rule with similar kind of examples is,

Pauls Online Notes : Calculus III - Chain Rule
Hello Sudharaka,
Thanks once again for the help and taking your time!:) That link seems really good I will read about it when I got time and I will be back if I got any question!:)

Regards,
\(\displaystyle |\pi\rangle\)
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Hello Sudharaka,
Thanks once again for the help and taking your time!:) That link seems really good I will read about it when I got time and I will be back if I got any question!:)

Regards,
\(\displaystyle |\pi\rangle\)
Glad to be of help. :)