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My friends argue this $d_t( \partial_{\dot{x}} g)=1+2\dot{\dot{x}} \not = \partial_t (\partial_{\dot{x} }g)$ where $g=t\dot x + x^2 + \dot{x}^2$. Why?

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What exactly do you mean by $d_t$ for a multivariable function? – gnometorule Feb 6 '13 at 13:39

Because they are derivatives of different functions.

Example $g(x,t)=t x^2$

When $x$ is itself a function of $t$, we have two options:

  • hold $x$ constant and take the partial derivative of $g(x,t)=tx^2$ with respect to $t$. This gives $\partial_t g = x^2$
  • treat $x$ as a function of $t$. Then we really look at the single-variable function $G(t):=g(x(t),t)$ but usually people don't bother introducing notation for it. The derivative is $G'(t)=x^2+t2x\dot x$.

You can treat the second computation as a partial derivative too; it's just that instead of holding $x$ constant, we hold an expression of $x,t$ constant. For example, if $x=t^3$, then we differentiate $g$ with respect to $t$ holding $x-t^3$ constant. This consideration occurs in mechanics when the change of coordinates is introduced.

Example with coordinate change

If your coordinates are $x,y$ and you decided to introduce a new coordinate $\tilde y=x+y$, then you should also introduce $\tilde x=x$ because the partial derivatives $\partial_x g$ and $\partial_{\tilde x}g$ will be different (even though $x$ and $\tilde x$ are the same thing). This takes a while to get used to.

Added later. The process of taking a partial derivative involves the following steps:

  1. Restrict the function to a curve
  2. Choose a parameter for that curve
  3. Differentiate the restricted function with respect to the chosen parameter.

For example, what is $\dfrac{\partial f}{\partial y}(1,2,3)$?

Step 1 is commonly expressed by saying "hold other variables constant". Here we take the partial derivative at $(1,2,3)$ "holding $x$ and $z$ constant", which means that we restrict the function to the line $x=1$, $z=3$. Notice that this step is not about the variable in which we will take the derivative.

Step 2: we choose the parameter for our line, namely $y$.

Step 3 is now unambigious: we differentiate a function of one variable.

But it does not have to be so rectangular all the time. Instead we can restrict $f$ to the curve formed by the intersection $z=x^2+y$, $x+y+z=6$. This means differentiation while holding the variables $u=z-x^2$ and $v=x+y+z$ constant. And our parameter could be $\tilde y = y$, or maybe $\tilde y = e^y-xyz$. The possibilities are infinite.

The traditional notation $\dfrac{\partial f}{\partial y}(1,2,3)$ hides step 1, taking for granted that the choice of restriction is obvious. This is the case in multivariable calculus, but often not the case in physics.

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Where can I find more examples about this? I find it misleading that sometimes I need to consider variables as constants and sometimes not, misleading. I cannot understand why this is allowed, possible to do some mistake when assuming variable as constant? – hhh Feb 11 '13 at 11:16
@hhh I added more explanation. I remember such discussions in analytical mechanics textbooks, but can't give a reference now. – user53153 Feb 11 '13 at 21:18
up vote 0 down vote accepted

$g=t\dot x+x^2+\dot x^2$

Total derivative

$\frac{dg}{dt}=\dot x+2x\frac{dx}{dt}+2\dot x \frac{d\dot x}{dt}$

Partial derivative where we consider other variables as constants

$\frac{\partial g}{\partial t}=\dot x$

Hence for all $t$ and for all $x$ $$\frac{dg}{dt}\not = \frac{\partial g}{\partial t}.$$

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