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Chain/partial proof

Petrus

Well-known member
Feb 21, 2013
739
Hello MHB,
anyone know good Link with good explaining on multivariable chain rule and partial derivate proof?I am having hard to understand the proof.
Regards,
\(\displaystyle |\pi\rangle\)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Hello MHB,
anyone know good Link with good explaining on multivariable chain rule and partial derivate proof?I am having hard to understand the proof.
Regards,
\(\displaystyle |\pi\rangle\)
Hi $|Petrus\rangle$,

Well, googling for it, this link seems to be as good as any.


Let me have a go at it to try to explain it.

Suppose we have a function $f(x,y)$.
And suppose both $x$ and $y$ are actually functions of $t$.
So we have $g(t)=f\big(x(t), y(t)\big)$.

Then the multivariable chain rule says that:
$$g'(t) = \frac d{dt}f\big(x(t), y(t)\big) = \frac{\partial}{\partial x}f\big(x(t), y(t)\big) \cdot \frac {d}{dt}x(t) + \frac{\partial}{\partial y}f\big(x(t), y(t)\big) \cdot \frac {d}{dt}y(t)$$
Or for short:
$$g'(t)=\frac {df}{dt} = \frac{\partial f}{\partial x} \cdot \frac {dx}{dt} + \frac{\partial f}{\partial y}\cdot \frac {dy}{dt}$$
 

Petrus

Well-known member
Feb 21, 2013
739
Hello I like Serena,
There is a another way i am suposed to learn I think, I am Really confused.. I know this pic is on Swedish but this is what I am suposed to understand


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

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Well, that first formula is the same as the first formula that I just gave you.
It's just with different symbols and a different notation.

In your slide they use $f_x'\big(g(t), h(t)\big)$ where I was using \(\displaystyle \frac{\partial}{\partial x}f\big(x(t), y(t)\big)\). They mean the same thing.
 

Petrus

Well-known member
Feb 21, 2013
739
Well, that first formula is the same as the first formula that I just gave you.
It's just with different symbols and a different notation.

In your slide they use $f_x'\big(g(t), h(t)\big)$ where I was using \(\displaystyle \frac{\partial}{\partial x}f\big(x(t), y(t)\big)\). They mean the same thing.
Hmm I start to confuse myself right now.. Well I am done for today and hopefully evrything start to make sense tomorow! Thanks for taking your time I like Serena!:) (I Will probably be Back tomorow, I just need to take some time with this..)
\(\displaystyle |\pi\rangle\)
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
To understand the proof of the multivariable chain rule, I think you will find it helpful to look first at an informal, nonrigorous demonstration, as given here (you need only look at the first page of that document). That demonstration uses the fact that if $z = f(x,y)$, and the variables $x, y$ are altered by small amounts $\Delta x,\,\Delta y$, then the corresponding change in $z$ is given by the approximate formula $\Delta z \approx \frac{\partial z}{\partial x}\Delta x + \frac{\partial z}{\partial y}\Delta y$. What the Swedish proof does is to take that informal approach and make it rigorous, replacing the approximate formula by an exact formula of the form $\Delta z = \frac{\partial z}{\partial x}\Delta x + \frac{\partial z}{\partial y}\Delta y + E(\Delta x, \Delta y)$. In that formula, the error term $E(\Delta x, \Delta y)$ represents the amount needed to convert the approximate formula into an exact equation. The essential fact about $E(\Delta x, \Delta y)$ is that it is small compared with $\Delta x$ and $\Delta y$. This is expressed by writing $E(\Delta x, \Delta y)$ as $E(\Delta x, \Delta y) = \rho(\Delta x, \Delta y)\sqrt{\Delta x^2 + \Delta y^2},$ where $\rho(\Delta x, \Delta y)$ is a function that tends to $0$ as $(\Delta x, \Delta y) \to (0,0).$

To sum up, make sure that you understand the ideas in the nonrigorous argument first, then go back to the rigorous approach and see what you can make of it.