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What is the second partial derivative $$ \frac{\partial^2 f(x)}{\partial y \partial z}, $$

where $x$ is a function $x = x(y, z)$. Is there a chain rule for this case? I can't find this anywhere, I can only find the rule for $\frac{\partial^2 f(x)}{\partial y^2}$.

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    $\begingroup$ Hint: $\frac{\partial^2 f(x)}{\partial y \partial z} = \frac{\partial}{\partial y}\left(\frac{\partial f(x)}{\partial z}\right) =\frac{\partial}{\partial z}\left(\frac{\partial f(x)}{\partial y}\right)$ $\endgroup$ – J. Bush Apr 21 '16 at 13:32
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You are given a function $t\mapsto f(t)$ of a single variable $t$ and a real-valued function $(y,z)\mapsto x(y,z)$ of the two variables $y$ and $z$. You then ask for the partial derivatives of the function $$\phi(y,z):=f\bigl(x(y,z)\bigr)\ .$$ The chain rule gives $$\phi_y=f'\>x_y,\qquad \phi_z=f'\>x_z$$ and then $$\phi_{yy}=f''\>(x_y)^2+f'\>x_{yy},\quad \phi_{yz}=f''\>x_z\>x_y+f'\>x_{yz},\quad \phi_{zz}=f''\>(x_z)^2+f'\>x_{zz}\ .$$

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