Analysis of Implicit Differentiation

[Math Amateur]

I am reading several books on multivariable analysis/calculus and am trying to get a precise and rigorous theoretical understanding of implicit differentiation, including the Implicit Function Theorem ... in particular I am reading:

Vector Calculus (Second Edition) by Susan Colley

and

Calculus: Volume II (Second Edition) by Tom Apostol

I am trying to carefully relate the working in Apostol's first worked example in Section 9.7, just following Section 9.7 on the derivative of a function defined implicitly ... ... Worked Example 1 reads as follows:
View attachment 4041As indicated above, I have tried to work through this example carefully relating it to Colley's definition of the Chain Rule and also trying to be explicit about the structure of the functions involved ... Colley's statement of the Chain Rule is as follows:
View attachment 4042
https://www.physicsforums.com/attachments/4043
I have also tried to follow (very carefully) Apostol's analysis in Section 9.6 which reads as follows:

https://www.physicsforums.com/attachments/4044
View attachment 4045

I will now provide my analysis and would very much appreciate someone critiquing my analysis and point out any shortcomings, misinterpretations or errors ...We have \(\displaystyle g(x,y) = 0\) and \(\displaystyle y = Y(x)\) for all \(\displaystyle x\) in some open interval \(\displaystyle (a,b)\) ... and we let

\(\displaystyle G(x) = g[ x, Y(x)]\) for \(\displaystyle x \in (a,b)\) ... ...

Then ...

Following Apostol, Section 9.6 we put

\(\displaystyle u_1(x,y) = x\)

and

\(\displaystyle u_2(x,y) = y = Y(x)\)Then let \(\displaystyle G(x) = g[ u_1(x,y), u_2(x,y) ]\)

Let \(\displaystyle \underline{u} (x,y) = ( u_1, u_2)\)

So we have the situation as shown below in Figure 1.
View attachment 4046

So continuing ... for \(\displaystyle x \in (a,b)\) we have\(\displaystyle D(g \circ \underline{u} ) (x,y) = Dg (u_1, u_2) D \underline{u} (x,y) \)

or

... if we let \(\displaystyle h = g \circ \underline{u}\) then we have:

\(\displaystyle Dh (x,y) = Dg (u_1, u_2) D \underline{u} (x,y) \)

... ...Now we have that ...\(\displaystyle Dh (x,y) = \begin {bmatrix} \frac{ \partial h }{ \partial x} & \frac{ \partial h }{ \partial y} \end{bmatrix}
\)\(\displaystyle Dg (u_1, u_2) = \begin {bmatrix} \frac{ \partial g }{ \partial u_1} & \frac{ \partial gh }{ \partial u_2} \end{bmatrix}\)

and

\(\displaystyle D \underline{u} (x,y) = \begin {bmatrix} \frac{ \partial u_1 }{ \partial x} & \frac{ \partial u_1 }{ \partial y} \\ \frac{\partial u_2 }{ \partial x} & \frac{ \partial u_2 }{ \partial y} \end{bmatrix}\)Now

\(\displaystyle Dh (x,y) = Dg (u_1, u_2) D \underline{u} (x,y) \)That is\(\displaystyle \begin {bmatrix} \frac{ \partial h }{ \partial x} & \frac{ \partial h }{ \partial y} \end{bmatrix} = \begin {bmatrix} \frac{ \partial g }{ \partial u_1} & \frac{ \partial gh }{ \partial u_2} \end{bmatrix} \begin {bmatrix} \frac{ \partial u_1 }{ \partial x} & \frac{ \partial u_1 }{ \partial y} \\ \frac{\partial u_2 }{ \partial x} & \frac{ \partial u_2 }{ \partial y} \end{bmatrix}\) ... ... ... (1)

So then (1) gives\(\displaystyle \frac{ \partial h }{ \partial x} = \frac{ \partial g }{ \partial u_1} \frac{ \partial u_1 }{ \partial x} + \frac{ \partial g }{ \partial u_2} \frac{ \partial u_2 }{ \partial x}\)But ... ... since \(\displaystyle u_1(x,y) = x\) we can view \(\displaystyle \frac{ \partial g }{ \partial u_1}\) as \(\displaystyle \frac{ \partial g }{ \partial x}\)and we also have \(\displaystyle \frac{ \partial u_1 }{ \partial x} = 1 \)Further ...

... since \(\displaystyle u_2(x,y) = y = Y(x)\) we can write \(\displaystyle \frac{ \partial g }{ \partial u_2}\) as \(\displaystyle \frac{ \partial g }{ \partial y}\)and we also have \(\displaystyle \frac{ \partial u_2 }{ \partial x} = \frac{ \partial Y(x) }{ \partial x} = \frac{dY}{dx} = Y'(x)\) since \(\displaystyle Y\) is a function of \(\displaystyle x\) ...so

\(\displaystyle \frac{ \partial h }{ \partial x} = \frac{ \partial g }{ \partial x} \cdot 1 + \frac{ \partial g }{ \partial y} \cdot Y'(x)\) But ...

\(\displaystyle \frac{ \partial h }{ \partial x} = \frac{ \partial g[ u_1(x,y), u_2(x,y) ] }{ \partial x} = G'(x)\)

so we can write \(\displaystyle G'(x) = \frac{ \partial g }{ \partial x} \cdot 1 + \frac{ \partial g }{ \partial y} \cdot Y'(x) \)

and so we have

\(\displaystyle Y'(x) = - \ \frac{ \frac{ \partial g }{ \partial x} }{ \frac{ \partial g }{ \partial y} }\)I would be most grateful if someone could critique my analysis above and point out any shortcomings, misinterpretations or errors ... or just confirm the above analysis is OK ...

Peter

 

[caffeinemachine]

I am reading several books on multivariable analysis/calculus and am trying to get a precise and rigorous theoretical understanding of implicit differentiation, including the Implicit Function Theorem ... in particular I am reading:

Vector Calculus (Second Edition) by Susan Colley

and

Calculus: Volume II (Second Edition) by Tom Apostol

I am trying to carefully relate the working in Apostol's first worked example in Section 9.7, just following Section 9.7 on the derivative of a function defined implicitly ... ... Worked Example 1 reads as follows:
As indicated above, I have tried to work through this example carefully relating it to Colley's definition of the Chain Rule and also trying to be explicit about the structure of the functions involved ... Colley's statement of the Chain Rule is as follows:I have also tried to follow (very carefully) Apostol's analysis in Section 9.6 which reads as follows:

I will now provide my analysis and would very much appreciate someone critiquing my analysis and point out any shortcomings, misinterpretations or errors ...We have \(\displaystyle g(x,y) = 0\) and \(\displaystyle y = Y(x)\) for all \(\displaystyle x\) in some open interval \(\displaystyle (a,b)\) ... and we let

\(\displaystyle G(x) = g[ x, Y(x)]\) for \(\displaystyle x \in (a,b)\) ... ...

Then ...

Following Apostol, Section 9.6 we put

\(\displaystyle u_1(x,y) = x\)

and

\(\displaystyle u_2(x,y) = y = Y(x)\)Then let \(\displaystyle G(x) = g[ u_1(x,y), u_2(x,y) ]\)

Let \(\displaystyle \underline{u} (x,y) = ( u_1, u_2)\)

So we have the situation as shown below in Figure 1.So continuing ... for \(\displaystyle x \in (a,b)\) we have\(\displaystyle D(g \circ \underline{u} ) (x,y) = Dg (u_1, u_2) D \underline{u} (x,y) \)

or

... if we let \(\displaystyle h = g \circ \underline{u}\) then we have:

\(\displaystyle Dh (x,y) = Dg (u_1, u_2) D \underline{u} (x,y) \)

... ...Now we have that ...\(\displaystyle Dh (x,y) = \begin {bmatrix} \frac{ \partial h }{ \partial x} & \frac{ \partial h }{ \partial y} \end{bmatrix}
\)\(\displaystyle Dg (u_1, u_2) = \begin {bmatrix} \frac{ \partial g }{ \partial u_1} & \frac{ \partial gh }{ \partial u_2} \end{bmatrix}\)

and

\(\displaystyle D \underline{u} (x,y) = \begin {bmatrix} \frac{ \partial u_1 }{ \partial x} & \frac{ \partial u_1 }{ \partial y} \\ \frac{\partial u_2 }{ \partial x} & \frac{ \partial u_2 }{ \partial y} \end{bmatrix}\)Now

\(\displaystyle Dh (x,y) = Dg (u_1, u_2) D \underline{u} (x,y) \)That is\(\displaystyle \begin {bmatrix} \frac{ \partial h }{ \partial x} & \frac{ \partial h }{ \partial y} \end{bmatrix} = \begin {bmatrix} \frac{ \partial g }{ \partial u_1} & \frac{ \partial gh }{ \partial u_2} \end{bmatrix} \begin {bmatrix} \frac{ \partial u_1 }{ \partial x} & \frac{ \partial u_1 }{ \partial y} \\ \frac{\partial u_2 }{ \partial x} & \frac{ \partial u_2 }{ \partial y} \end{bmatrix}\) ... ... ... (1)

So then (1) gives\(\displaystyle \frac{ \partial h }{ \partial x} = \frac{ \partial g }{ \partial u_1} \frac{ \partial u_1 }{ \partial x} + \frac{ \partial g }{ \partial u_2} \frac{ \partial u_2 }{ \partial x}\)But ... ... since \(\displaystyle u_1(x,y) = x\) we can view \(\displaystyle \frac{ \partial g }{ \partial u_1}\) as \(\displaystyle \frac{ \partial g }{ \partial x}\)and we also have \(\displaystyle \frac{ \partial u_1 }{ \partial x} = 1 \)Further ...

... since \(\displaystyle u_2(x,y) = y = Y(x)\) we can write \(\displaystyle \frac{ \partial g }{ \partial u_2}\) as \(\displaystyle \frac{ \partial g }{ \partial y}\)and we also have \(\displaystyle \frac{ \partial u_2 }{ \partial x} = \frac{ \partial Y(x) }{ \partial x} = \frac{dY}{dx} = Y'(x)\) since \(\displaystyle Y\) is a function of \(\displaystyle x\) ...so

\(\displaystyle \frac{ \partial h }{ \partial x} = \frac{ \partial g }{ \partial x} \cdot 1 + \frac{ \partial g }{ \partial y} \cdot Y'(x)\) But ...

\(\displaystyle \frac{ \partial h }{ \partial x} = \frac{ \partial g[ u_1(x,y), u_2(x,y) ] }{ \partial x} = G'(x)\)

so we can write \(\displaystyle G'(x) = \frac{ \partial g }{ \partial x} \cdot 1 + \frac{ \partial g }{ \partial y} \cdot Y'(x) \)

and so we have

\(\displaystyle Y'(x) = - \ \frac{ \frac{ \partial g }{ \partial x} }{ \frac{ \partial g }{ \partial y} }\)I would be most grateful if someone could critique my analysis above and point out any shortcomings, misinterpretations or errors ... or just confirm the above analysis is OK ...

Peter

Hello Peter,

I would like to help you but your post is too long to read.

I suggest you do the following.

Write an expository post on your understanding of the implicit differentiation using references from textbooks as little as possible.

Thus, each notation you use will need to be defined within your post, except, of course, the standard ones.

Then I think it would be more amenable to people.

If you choose to do so, drop me a visitor message.

I will try to help.

 

[Math Amateur]

Hello Peter,

I would like to help you but your post is too long to read.

I suggest you do the following.

Write an expository post on your understanding of the implicit differentiation using references from textbooks as little as possible.

Thus, each notation you use will need to be defined within your post, except, of course, the standard ones.

Then I think it would be more amenable to people.

If you choose to do so, drop me a visitor message.

I will try to help.

Hi caffeinemachine,

Thanks for being willing to help ... appreciate it ...

I have been reflecting on what you have said about the length of the post ... but ... I do not think I can express it in a shorter post ...

Note a couple of points ...

1. The uploads are only needed for a reader to check the notation and to get an idea of the context ... they can be skimmed or even ignored by the reader with a good knowledge of implicit differentiation (which I suspect will include a number of MHB members) ... so a number of readers will only have to look at what I typed ...

2. I am only asking for someone to read what I have typed/written and confirm it is OK ...

Thanks again ...

Peter