Chain Rule & Partial Derivative Proofs: Understand Easily!

[Petrus]

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,
$$|\pi\rangle$$

 

[I like Serena]

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,
$$|\pi\rangle$$

Hi $|Petrus\rangle$,

Well, googling for it, http://www.math.hmc.edu/calculus/tutorials/multichainrule/ 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]

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,
$$|\pi\rangle$$

 

[I like Serena]

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 $$\frac{\partial}{\partial x}f\big(x(t), y(t)\big)$$. They mean the same thing.

 

[Petrus]

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 $$\frac{\partial}{\partial x}f\big(x(t), y(t)\big)$$. 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..)
$$|\pi\rangle$$

 

[Opalg]

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 http://math.ucsd.edu/~wgarner/reference/math10c_su10/lectures/chain_rule.pdf (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.