Differentiate with respect to one variable while treating the others as constants for example:
$$ \begin{aligned} f(x,y)&=x^2y \\ \frac{\partial f}{\partial x}=2xy\quad&\text{and}\quad\frac{\partial f}{\partial y}=x^{2} \end{aligned} $$
symbol: $\partial$ (squigly d)
$$ f_{x}\equiv\frac{\partial f}{\partial x} $$
variable constant can be written:
$$ \left(\frac{\partial f}{\partial x}\right)_y $$
seccond order:
$$ \frac{\partial^{2}f}{\partial x^{2}}\quad\textrm{or}\quad f_{xx} $$
mixed second order:
$$ \frac{\partial^{2}f}{\partial x\partial y}=\frac{\partial^{2}f}{\partial y\partial x} $$
$$ \frac{\mathrm dz}{\mathrm dt}=\frac{\partial z}{\partial x}\frac{\mathrm dx}{\mathrm dt} +\frac{\partial z}{\partial y}\frac{\mathrm dy}{\mathrm dt} $$
if $f(x,y)$ and $x(s,t)\; ;\,y(s,t)$ then:
$$ \frac{\partial f}{\partial s}= \frac{\partial f}{\partial x}\frac{\partial x}{\partial s}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial s} \; \text{ and }\;\frac{\partial f}{\partial t}= \frac{\partial f}{\partial x}\frac{\partial x}{\partial t}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial t} $$
if $f(x,y)$ then $x(r,\theta)=r\cos\theta$ and $y(r,\theta)=r\sin\theta$
we can convert $f(x,y)$ → $g(r,\theta)$ using: