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I have learned for the total derivative: $$\frac{dF(x(z),y(z))}{dz} = \frac{\partial F(x,y)}{\partial x} \frac{dx}{dz} + \frac{\partial F(x,y)}{\partial y} \frac{dy}{dz}.$$

Now I need the partial derivative: $$ \frac{\partial F(\phi_1(x,z),\phi_2(y,z))}{\partial z} .$$

Is it correct that: $$ \frac{\partial F(\phi_1(x,z),\phi_2(y,z))}{\partial z} = \frac{\partial F(\phi_1,\phi_2)}{\partial \phi_1} \frac{\partial \phi_1}{\partial z} + \frac{\partial F(\phi_1,\phi_2)}{\partial \phi_2} \frac{\partial \phi_2}{\partial z} ?$$

I am confused because of the difference between total and partial derivative.

BLAZE
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Gerard
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  • if $x$ and $y$ are independent from variable $z$ then your result is correct. Otherwise it is not correct. – math Mar 13 '15 at 08:55

1 Answers1

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If $x$ and $y$ are independent from variable z then your result is correct.

If $x=x(z)$ and $y=y(z)$, then it should be

$$ \frac{\partial F(\phi_1(x,z),\phi_2(y,z))}{\partial z} = \frac{\partial F}{\partial \phi_1} \left( \frac{\partial \phi_1}{\partial x}\frac{\partial x}{\partial z} + \frac{\partial \phi_1}{\partial z} \right) + \frac{\partial F}{\partial \phi_2} \left( \frac{\partial \phi_1}{\partial y}\frac{\partial y}{\partial z}+ \frac{\partial \phi_2}{\partial z} \right) $$

Ben Grossmann
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math
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    You could explicitly state that if $x$ and $y$ are independent of $z$, their derivatives with respect to $z$ vanish. (OP's result) – Demosthene Mar 13 '15 at 09:50