Partial derivative

  • Thread starter Niles
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  • #1
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Homework Statement


Hi

Say I have a function [itex]f(x(t), t)[/itex]. I am not 100% sure of the difference between
[tex]
\frac{df}{dt}
[/tex]
and
[tex]
\frac{\partial f}{\partial t}
[/tex]
Is it correct that the relation between these two is (from the chain rule)
[tex]
\frac{df}{dt} = \frac{\partial f}{\partial t} + \frac{\partial f}{\partial x}\frac{dx}{dt}
[/tex]
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement


Hi

Say I have a function [itex]f(x(t), t)[/itex]. I am not 100% sure of the difference between
[tex]
\frac{df}{dt}
[/tex]
and
[tex]
\frac{\partial f}{\partial t}
[/tex]
Is it correct that the relation between these two is (from the chain rule)
[tex]
\frac{df}{dt} = \frac{\partial f}{\partial t} + \frac{\partial f}{\partial x}\frac{dx}{dt}
[/tex]

It is easy to be confused by the ambiguity of ##\frac{\partial f}{\partial t}## symbol. If you write the expression instead as ##f(u,v)## where ##u = x(t),~v=t## you would write$$
\frac{df}{dt} = f_u\frac {du}{dt} + f_v\frac{dv}{dt}=f_u\frac{dx}{dt}+f_v\cdot 1$$You wouldn't normally talk about ##\frac{\partial f}{\partial t}## as though ##f## depended on another variable also. But as the chain rule gives, you need the partials of ##f## with respect to each of its arguments. If you understand that ##\frac{\partial f}{\partial x}## and ##\frac{\partial f}{\partial t}## in this setting mean the partials of ##f## with respect to its first and second arguments, you should be OK.
 

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