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Anonymous
2016-01-25 08:15:02 Post No. 7809153
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Anonymous
2016-01-25 08:15:02
Post No. 7809153
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case 1)
f = f(x,y,z) where we know that x,y,z are INDEPENDENT of each other
- the total derivative:
[math]\frac {df} {dx} = \frac {\partial f} {\partial x} \frac {dx} {dx} + \frac {\partial f} {\partial y} \frac {dy} {dx} + \frac {\partial f} {\partial z} \frac {dz} {dx}[/math]
[math]\frac {df} {dx} = \frac {\partial f} {\partial x} (1) + \frac {\partial f} {\partial y} (0) + \frac {\partial f} {\partial z} (0)[/math]
[math]\frac {df} {dx} = \frac {\partial f} {\partial x}[/math]
- the partial:
[math]\frac {\partial f} {\partial x} = \frac {\partial f} {\partial x}[/math]
case 2)
f = f(x,y,z) where we know that x,y,z are DEPENDENT on x
- the total derivative:
[math]\frac {df} {dx} = \frac {\partial f} {\partial x} \frac {dx} {dx} + \frac {\partial f} {\partial y} \frac {dy} {dx} + \frac {\partial f} {\partial z} \frac {dz} {dx}[/math]
[math]\frac {df} {dx} = \frac {\partial f} {\partial x} (1) + \frac {\partial f} {\partial y} \frac {dy} {dx} + \frac {\partial f} {\partial z} \frac {dz} {dx}[/math]
[math]\frac {df} {dx} = \frac {\partial f} {\partial x} + \frac {\partial f} {\partial y} \frac {dy} {dx} + \frac {\partial f} {\partial z} \frac {dz} {dx}[/math]
- the partial:
[math]\frac {\partial f} {\partial x} = \frac {\partial f} {\partial x} + \frac {\partial f} {\partial y} \frac {\partial y} {\partial x} + \frac {\partial f} {\partial z} \frac {\partial z} {\partial x}[/math]
mfw partials and totals are same and mathfags just like to jerk themselves off with different notations
