The chain rule (General version)

In summary: Keep learning and good luck with your studies! In summary, the conversation is about solving a problem involving a multivariable function using the chain rule. The problem is solved using the given equation and the chain rule. The conversation also includes a reference for further understanding of the chain rule.
  • #1
Petrus
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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 want to be specefic. I want to 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,
 

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  • #2
Petrus said:
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 want to be specefic. I want to 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
 
  • #3
Sudharaka said:
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,
\(\displaystyle |\pi\rangle\)
 
  • #4
Petrus said:
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. :)
 

FAQ: The chain rule (General version)

What is the chain rule in calculus?

The chain rule is a rule in calculus that is used to find the derivative of a composite function. It allows us to find the rate of change of a function that is composed of two or more functions.

How is the chain rule used?

The chain rule is used by taking the derivative of the outer function and multiplying it by the derivative of the inner function. This is represented mathematically as (f(g(x)))' = f'(g(x)) * g'(x).

Why is the chain rule important?

The chain rule is important because it is used to find the derivatives of many functions that are used in real-world applications. It also allows us to find the rate of change of a function that is composed of multiple functions, which is often necessary in physics, engineering, and other fields.

Can the chain rule be applied to any composite function?

Yes, the chain rule can be applied to any composite function, as long as the functions within the composition are differentiable. This means that the functions must have a well-defined derivative at each point in their domain.

What are some common mistakes when using the chain rule?

Some common mistakes when using the chain rule include forgetting to multiply by the derivative of the inner function, incorrect use of the power rule, and not properly simplifying the final expression. It is important to carefully apply the chain rule and check for errors when finding derivatives of composite functions.

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