1. May 29, 2012

### Hassan2

Dear all,

I have a confusion about partial derivatives.

Say I have a function as

$y=f(x,t)$

and we know that
$x=g(t)$

1. Does it make sense to talk about partial derivatives like $\frac{\partial y}{\partial x}$ and $\frac{\partial y}{\partial t}$ ?

I doubt, because the definition of partial derivative is the change in the function due to the change on the selected variable ( other variables are kept constant).

2. If it does not make sense, then how in Euler–Lagrange equation we use the partial derivatives with respect to a function(x(t)) and its derivative(x'(t)) while they depend on one another.

2. May 29, 2012

### chiro

Hey Hassan2.

For 1. Yes it makes sense to do both but you need to factor in things like the chain rule for this particular example: as long as you are taking into account these kinds of factors, then yes it's ok. Your partial with respect to t can take into account your g(t) using the chain rule.

If the function is differentiable in the the region you are considering, the derivative will make sense: it's guaranteed to as a consequence of differentiability holding.

3. May 29, 2012

### Hassan2

Thanks Chiro,

Yes, I checked the derivation of Euler-Largrange equation once again, and i found that the derivatives wit respect to x and x' appear as a result of applying chain rule.

Thanks again.

4. May 29, 2012

### HallsofIvy

Staff Emeritus
The notation $\partial f/\partial x$ means the derivative of f with respect to x while holding x constant- and ignoring the fact that x is a function of t. We are really dealing with the "form" of f rather than the content.

The same is true of $\partial f/\partial t$. However, we can, using x= g(t), think of f as a function of t only- f(x, t)= f(g(t), t). In that case, by the chain rule,
$$\frac{df}{dt}= \frac{\partial f}{\partial t}+ \frac{\partial f}{\partial x}\frac{dx}{dt}$$

Example: if $f(x,t)= 3tx^2+ e^x$ then $\partial f/\partial x= 6tx+ e^x$ and $\partial f/\partial t= 3x^2$. That has nothing to do with x being a function of t or vice-versa.

But if we also know that $x= g(t)= 2t^2+ t$ we can write $f(t)= 3t(2t^2+ t)+ e^{2t^2+ t}= 12t^5+ 12t^4+ 3t^3+ e^{2t^2+ t}$ so that the derivative is
$$\frac{df}{dt}= 60t^4+ 48t^3+ 9t^2+ (4t+ 1)e^{2t^2+ t}$$

Or, you could use the chain rule as I said:
$\partial f/\partial t= 3x^2$ and $\partial f/\partial x= 3tx+ e^x$, as above, while $dx/dt= 4t+1$ and so
$$\frac{df}{dx}= 3x^2+ (3tx+ e^x)(4t+ 1)$$
and, replacing x in that with $2t^2+ t$
$$\frac{df}{dx}= 3(2t^2+ t)+ (3t(2t^2+ t)+ e^{2t^2+ t})(4t+1)$$
gives the same thing.

5. May 29, 2012

### algebrat

Strictly speaking, one should not write f(t)=f(g(t),t). The function on the left should have gotten a new name. That is the source of some confusion.

6. May 29, 2012

### arildno

Agreed!
What we should write, is something like:

F(t)=f(g(t),t)