- Thread starter
- #1

#### Petrus

##### Well-known member

- Feb 21, 2013

- 739

anyone know good Link with good explaining on multivariable chain rule and partial derivate proof?I am having hard to understand the proof.

Regards,

\(\displaystyle |\pi\rangle\)

- Thread starter Petrus
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- Thread starter
- #1

- Feb 21, 2013

- 739

anyone know good Link with good explaining on multivariable chain rule and partial derivate proof?I am having hard to understand the proof.

Regards,

\(\displaystyle |\pi\rangle\)

- Admin
- #2

- Mar 5, 2012

- 9,773

Hi $|Petrus\rangle$,

anyone know good Link with good explaining on multivariable chain rule and partial derivate proof?I am having hard to understand the proof.

Regards,

\(\displaystyle |\pi\rangle\)

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:

$$g'(t)=\frac {df}{dt} = \frac{\partial f}{\partial x} \cdot \frac {dx}{dt} + \frac{\partial f}{\partial y}\cdot \frac {dy}{dt}$$

- Thread starter
- #3

- Feb 21, 2013

- 739

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,

\(\displaystyle |\pi\rangle\)

- Admin
- #4

- Mar 5, 2012

- 9,773

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.

- Thread starter
- #5

- Feb 21, 2013

- 739

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..)

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.

\(\displaystyle |\pi\rangle\)

- Moderator
- #6

- Feb 7, 2012

- 2,807

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.