Partial Differential: For f(x,t), df = [(partial f)/(partial x)] * dx

Viona

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Nov 22, 2018
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Hello,

In a book I found that for f(x,t):

dff.png

But I could not remember this rule of partial differentiation. Any help?

Thanks
 
For a general function of two variables, that is wrong. The definition of "differential" of a function of one variable is \(\displaystyle df= \frac{df}{dx}dx\). But for a function of two variables, x and t, \(\displaystyle df= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial t}dt\).

Was this function, f, assumed to have any special properties?
 
For a general function of two variables, that is wrong. The definition of "differential" of a function of one variable is \(\displaystyle df= \frac{df}{dx}dx\). But for a function of two variables, x and t, \(\displaystyle df= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial t}dt\).

Was this function, f, assumed to have any special properties?
Halls

Viona is reading some book in physics that seems to be more than a bit casual in its use of mathematical notation. She really needs to give more context for us to be of any assistance.
 
Halls

Viona is reading some book in physics that seems to be more than a bit casual in its use of mathematical notation. She really needs to give more context for us to be of any assistance.
Ew! Not good even for a Physics book. Unless it's supposed to be something like \(\displaystyle df = \dfrac{\partial f}{\partial x^i} dx^i\) and we're using the Einstein convention.

@Viona: If this is a Physics book that you have to use then post here liberally. Otherwise throw it in the fire. Any text that would be this lax needs to be euthanized.

-Dan
 
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