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case 1)
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You are currently reading a thread in /sci/ - Science & Math

case 1)
f = f(x,y,z) where we know that x,y,z are INDEPENDENT of each other

- the total derivative:
$\frac {df} {dx} = \frac {\partial f} {\partial x} \frac {dx} {dx} + \frac {\partial f} {\partial y} \frac {dy} {dx} + \frac {\partial f} {\partial z} \frac {dz} {dx}$
$\frac {df} {dx} = \frac {\partial f} {\partial x} (1) + \frac {\partial f} {\partial y} (0) + \frac {\partial f} {\partial z} (0)$
$\frac {df} {dx} = \frac {\partial f} {\partial x}$

- the partial:
$\frac {\partial f} {\partial x} = \frac {\partial f} {\partial x}$

case 2)
f = f(x,y,z) where we know that x,y,z are DEPENDENT on x

- the total derivative:
$\frac {df} {dx} = \frac {\partial f} {\partial x} \frac {dx} {dx} + \frac {\partial f} {\partial y} \frac {dy} {dx} + \frac {\partial f} {\partial z} \frac {dz} {dx}$
$\frac {df} {dx} = \frac {\partial f} {\partial x} (1) + \frac {\partial f} {\partial y} \frac {dy} {dx} + \frac {\partial f} {\partial z} \frac {dz} {dx}$
$\frac {df} {dx} = \frac {\partial f} {\partial x} + \frac {\partial f} {\partial y} \frac {dy} {dx} + \frac {\partial f} {\partial z} \frac {dz} {dx}$

- the partial:
$\frac {\partial f} {\partial x} = \frac {\partial f} {\partial x} + \frac {\partial f} {\partial y} \frac {\partial y} {\partial x} + \frac {\partial f} {\partial z} \frac {\partial z} {\partial x}$

mfw partials and totals are same and mathfags just like to jerk themselves off with different notations
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>mfw sci doesnt reply to anything that doesnt have

.9999999 = 1???
earth is flat faggots
meme musk rockets
le fucking ai doom bots
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>>7809153
What the fuck are you doing? The total derivative is as you wrote it. The partial derivative does derivation assuming the other variables constant.

That is \frac{\partial f}{\partial x} = \frac{\partial f}{\partial x} on both occasions.
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>>7809337
Apparently I must do something more for latex output to appear.