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+[[Calculus 3]]
+- Chain rule
+	- If there is a function $z(x(t), y(t))$ then $\frac{dz}{dt} = \frac{dz}{dx} \frac{dx}{dt} + \frac{dz}{dy} \frac{dy}{dt}$
+	- If $z=f(x(u,v), y(u,v))$ then $$\frac{{\partial z}}{\partial u} = \frac{{\partial z}}{\partial x} \frac{{\partial x}}{\partial u} + \frac{{\partial z}}{\partial y} \frac{{\partial y}}{\partial u}$$ $$\frac{{\partial z}}{\partial v} = \frac{{\partial z}}{\partial x} \frac{{\partial x}}{\partial v} + \frac{{\partial z}}{\partial y} \frac{{\partial y}}{\partial v}$$ 
+- Implicit differentiation
+	- If $z$ is defined implicitly as a function of $x$ and $y$, then $$ \frac{dz}{dx} = - \frac{{\frac{{\partial f}}{\partial x}}}{\frac{{\partial f}}{\partial z}}$$ $$ \frac{dz}{dx} = - \frac{{\frac{{\partial f}}{\partial y}}}{\frac{{\partial f}}{\partial z}}$$
+- Critical points
+	- For functions of two variables, this is when they both equal 0 or when one is undefined
+	- Second derivative test: $$D = f_{x x}(x_{0}, y_{0})f_{y y}(x_{0}, y_{0}) - (f_{x y}(x_{0}, y_{0}))^2$$
+		- if $D>0$ and $f_{x x}(x_{0}, y_{0})>0$ then $f$ has a local minimum at $(x_{0}, y_{0})$
+		- if $D>0$ and $f_{x x}(x_{0}, y_{0})<0$ then $f$ has a local maximum at $(x_{0}, y_{0})$
+		- if $D<0$ then $f$ has a saddle point at $(x_{0}, y_{0})$
+		- if $D=0$ then the test is inconclusive