Confusion about partial derivatives

[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.

Your help would be appreciated.

 

[chiro]

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.

Your help would be appreciated.

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.

 

[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.

 

[HallsofIvy]

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.

 

[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.

 

[arildno]

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.

Agreed!
What we should write, is something like:

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