Definition: Function of two variables
A real-valued function \(f\) defined on a subset \(\mathcal{D}\) of \(\mathbb{R}^2\) is a rule that assigns to each point \((x,y)\) in \(\mathcal{D}\) a real number \(f(x,y)\). The set \(\mathcal{D}\) is called the domain of \(f\), and the range \(\mathcal{R}\) of \(f\) is the set of all real numbers \(f(x,y)\) as \((x,y)\) varies over the domain \(\mathcal{D}\).
Given a function \(\displaystyle z = f(x,y)\) with domain \(\mathcal{D}\). A vertical trace can be
either the set of points satisfying the equation \(f(a,y) = z\) for a given constant \(a\)
or the set of points satisfying the equation \(f(x,b) = z\) for a given constant \(b\)
Definition: Function of several variables
A real-valued function \(f\) defined on a subset \(\mathcal{D}\) of \(\mathbb{R}^n\) is a rule that assigns to each point \((x_1,\dots,x_n)\) in \(\mathcal{D}\) a real number \(f(x_1,\dots,x_n)\). The set \(\mathcal{D}\) is called the domain of \(f\), and the range \(\mathcal{R}\) of \(f\) is the set of all real numbers \(f(x_1,\dots,x_n)\) as \((x_1,\dots,x_n)\) varies over the domain \(\mathcal{D}\).
Let $ \displaystyle\lim_{(x,y)\to(a,b)}f(x,y) = F $ and $ \displaystyle \lim_{(x,y)\to(a,b)}g(x,y) =G $ then
$$\begin{array}{rcl} \displaystyle\lim_{(x,y)\to(a,b)}\left(\alpha f(x,y) \pm \beta g(x,y) \right) &=&\alpha F\pm \beta G \\[2ex] \displaystyle\lim_{(x,y)\to(a,b)}f(x,y)g(x,y) &=&FG \end{array} $$Let $ \displaystyle\lim_{(x,y)\to(a,b)}f(x,y) = F $ and $ \displaystyle \lim_{(x,y)\to(a,b)}g(x,y) =G \ne 0 $ then
$$\begin{array}{rcl} \displaystyle\lim_{(x,y)\to(a,b)}\frac{f(x,y)}{g(x,y)} & = & \displaystyle\frac{F}{G} \end{array} $$if $\displaystyle \lim_{(x,y)\to(a,b)}f(x,y) = F $ and $n$ is positive integer then
$$\lim_{(x,y)\to(a,b)}{\left(f(x,y)\right)}^{n} =F^n $$Let $\displaystyle \lim_{(x,y)\to(a,b)}f(x,y) = F $ and $n$ is positive odd integer then
$$\lim_{(x,y)\to(a,b)}\sqrt[n]{\left(f(x,y)\right)} =\sqrt[n]{F} $$if $n$ is even and $F\ge 0 $ the above equation also holds.
Definition: Partial derivative with respect to $x$
Let $f:{\mathbb{R}}^2\to{\mathbb{R}}$ be a function of two variables $z= f(x,y)$. The partial derivative of $f$ with respect to $x$ is $$\begin{array}{rcl} f_x = \frac{\partial}{\partial x}\, f =\frac{\partial\,f}{\partial x} &=&\displaystyle \lim_{h\to0} \frac{f(x+h,y)-f(x,y)}{h} \end{array}$$
Definition: Partial derivative with respect to $y$
Let $f:{\mathbb{R}}^2\to{\mathbb{R}}$ be a function of two variables $z= f(x,y)$. The partial derivative of $f$ with respect to $y$ is $$\begin{array}{rcl} f_x = \frac{\partial}{\partial x}\, f =\frac{\partial\,f}{\partial x} &=&\displaystyle \lim_{h\to0} \frac{f(x,y+h)-f(x,y)}{h} \end{array}$$
Definition: Partial derivative with respect to $x_i$
Let $f:{\mathbb{R}}^n\to{\mathbb{R}}$ be a function of $n$ variables $ f(x_1,\dots,x_n)$. The partial derivative of $f$ with respect to $x_i$ is $$\begin{array}{rcl} f_{x_i}&=& = \frac{\partial}{\partial x_i}\, f =\frac{\partial\,f}{\partial x_i} \\\\&=&\displaystyle \lim_{h\to0} \frac{f(x_1,\dots,x_{i-1},x_i+h,x_{i+1},\dots,x_i)- f(x_1,\dots,x_i,\dots,x_i)}{h} \end{array}$$
Definition: Tangent plane equation
Let $S$ be a surface defined by a function $z=f(x,y)$ and let $\displaystyle P = \left[\begin{array}{r}P_x\\P_y\end{array}\right] $ be a point in the domain of $f$. Then the equation of the tangent plane to $S$ at $P$ is
$$ z = f(P_x,P_y) + f_x(P_x,P_y)(x-P_x)+ f_y(P_x,P_y)(y-P_y) $$Definition: Differentiable function
A function $f(x,y)$ is differentiable at $\displaystyle P = \left[\begin{array}{r}P_x\\P_y\end{array}\right] $ if for all pairs $(x,y)$ contained in a ball centered at $P$ we can write
$$ f(x,y) = f(P_x,P_y) + f_x(P_x,P_y)(x-P_x) + f_y(P_x,P_y)(y-P_y)+ E(x,y) $$where the error term $E(x,y)$ satisfies
$$ \lim_{(x,y)\to(P_x,P_y)} \frac{E(x,y)} {\sqrt{{\left(x-P_x\right)}^2+{\left(y-P_y\right)}^2}} =0 $$Let $z=f(x,y)$ and $\displaystyle P = \left[\begin{array}{r}P_x\\P_y\end{array}\right] \in\mathcal{D}_f$. Let $\displaystyle \Delta x$ and $\displaystyle \Delta y$ be such that $\displaystyle \left[\begin{array}{r}P_x+\Delta x\\P_y+\Delta y\end{array}\right] \in\mathcal{D}_f$. If $f$ is differentiable then the differentials $\mathrm{d}\,{x}$ and $\mathrm{d}\,{y}$ are defined as
$$\mathrm{d}\,{x} = \Delta x\qquad \mathrm{d}\,{y} = \Delta y$$the total differential $\mathrm{d}\,{z}$ is
$$\mathrm{d}\,{z} = f_x(P)\mathrm{d}\,{x} + f_y(P)\mathrm{d}\,{y}$$Definition: Differentiable function
A function $f(x_1,\dots,x_n)$ is differentiable at $\displaystyle P = \left[\begin{array}{c}P_{x_1}\\\vdots\\P_{x_n}\end{array}\right] $ if for all values $(\hat{x}_1,\dots,\hat{x}_n)$ contained in a ball centered at $P$ we can write
$$ f(\hat{x}_1,\dots,\hat{x}_n) = f(P) + \sum_{i=1}^{n}f_{x_i}(P)(\hat{x}_i-P_{x_i}) + E(\hat{x}_1,\dots,\hat{x}_n) $$where the error term $E(x_1,\dots,x_n)$ satisfies
$$ \lim_{(x_1,\dots,x_n)\to P} \frac{E(x_1,\dots,x_n)}{\sqrt{\displaystyle\sum_{i=1}^{n}{\left(x_{i}-P_{x_i}\right)}^2}} =0 $$Theorem: One variable chain rule
Let $z = f(x,y)$ be differentiable function and $x=x(t)$ and $y=y(t)$. Then $z$ is a differentiable function of $t$ and
$$ \dfrac{\mathrm{d}\,{z}}{\mathrm{d}\,{t}} = \dfrac{\partial z}{\partial {x}}\dfrac{\mathrm{d}\,{x}}{\mathrm{d}\,{t}} +\dfrac{\partial z}{\partial {y}}\dfrac{\mathrm{d}\,{y}}{\mathrm{d}\,{t}} $$Let $z = f(x_1,x_2,\dots,x_n)$ be differentiable function and for all $i\in[1,\dots,n]$ we have that $x_i=x_i(t_1,t_2,\dots, t_m)$. Then
$$ \dfrac{\partial {z}}{\mathrm{d}\,{t_j}} = \dfrac{\partial z}{\partial{x_1}}\dfrac{\partial{x_1}}{\partial{t_j}} + \dfrac{\partial z}{\partial{x_2}}\dfrac{\partial{x_2}}{\partial{t_j}} + \cdots + \dfrac{\partial z}{\partial{x_n}}\dfrac{\partial{x_n}}{\partial{t_j}} $$Theorem: Implicit differentiation
Suppose $\displaystyle z = f(x,y) $ defines $y$ implicitly as a function $y=g(x)$ of $x$ via the equation $f(x,y) = 0$. Then
$$ \begin{array}{rcl} \displaystyle \dfrac{\mathrm{d}\,{y}}{\mathrm{d}\,{x}} &=& -\dfrac{\dfrac{\partial f}{\partial x}}{\dfrac{\partial f}{\partial y}} \end{array} $$ provided that $$ \begin{array}{rcl} \displaystyle \displaystyle {\dfrac{\partial f}{\partial y}} & \ne & 0 \end{array} $$Theorem: Implicit differentiation
Suppose $\displaystyle 0 = f(x,y,z) $ defines $z$ implicitly as a differentiable function in $x$ and $y$ Then
$$ \begin{array}{rcl} \displaystyle \dfrac{\partial{z}}{\partial {x}} = -\dfrac{\dfrac{\partial f}{\partial x}}{\dfrac{\partial f}{\partial z}} &\qquad& \dfrac{\partial{z}}{\partial {x}} = -\dfrac{\dfrac{\partial f}{\partial x}}{\dfrac{\partial f}{\partial z}} \end{array} $$ provided that $$ \begin{array}{rcl} \displaystyle {\dfrac{\partial f}{\partial z}} & \ne & 0 \end{array} $$Definition: Directional derivative
Let $f(x,y):{\mathbb{R}}^2\to{\mathbb{R}}$ with domain ${\mathcal{D}_f}$ and let $(a,b)\in{\mathcal{D}_f}$. Let $\vec{u}$ be a unit vector. The directional derivative of ${f}$ at $(a,b)$ in the direction of $\vec{u}$ is
$$\begin{array}{rcl} {\mathrm{D}}_{\vec{u}}\,f(a,b) &=& \lim_{h \to 0} \dfrac{f((a,b) + h\vec{u}) - f(a,b)}{h} \\\\ &=& \lim_{h \to 0} \dfrac{f((a+hu_x,b+hu_y)) - f(a,b)}{h} \\\\ &=& \lim_{h \to 0} \dfrac{f((a+h\cos\theta,b+h\sin\theta)) - f(a,b)}{h} \end{array}$$
if the limit exists
Theorem: Directional derivative
Let $f(x,y):{\mathbb{R}}^2\to{\mathbb{R}}$ with domain ${\mathcal{D}_f}$. Let $\vec{u}=\cos\theta \vec{i}+\sin\theta \vec{j}$. Assume that $f_x$ and $f_y$ exist. Then
$$\begin{array}{rcl} {\mathrm{D}}_{\vec{u}}\,f(a,b) &=& f_x(x,y)\cos\theta +f_y(x,y)\sin\theta \\\\ &=& \left(\begin{array}{c}f_x\\f_y\end{array}\right) \cdot \left(\begin{array}{c}\cos\theta\\\sin\theta\end{array}\right) \end{array}$$Theorem: Properties of directional derivative
Let $f(x,y):{\mathbb{R}}^2\to{\mathbb{R}}$ be differentiable at $(a,b)\in{\mathcal{D}_f}$. Then
Let $f(x,y):{\mathbb{R}}^2\to{\mathbb{R}}$ on domain ${\mathcal{D}}_{f}$. A point $(a,b)\in{\mathcal{D}}_{f}$ is critical points for for $f(x,y)$ if one of the following holds
Theorem: Second derivative test
Let $f(x,y):{\mathbb{R}}^2\to{\mathbb{R}}$ on domain ${\mathcal{D}}$ and let $(a,b)\in{\mathcal{D}}$. If $f_x,f_y,f_{xx},f_{yy},f_{xy}$ and $f_{yx}$ are continuous on a some ball containing $(a,b)$ then let
$$D =\det \left(\displaystyle \begin{array}{rr} f_{xx}(a,b)&f_{yx}(a,b) \\ f_{yx}(a,b)&f_{yy}(a,b) \end{array} \right) = f_{xx}(a,b)f_{yy}(a,b) - {\left(f_{xy}(a,b)\right)}^2 $$Let $f(x,y)$ and $g(x,y)$ be smooth functions, and suppose that $c$ is a scalar constant such that $\nabla g(x,y) \ne \vec{0}$ for all $(x,y)$ that satisfy the equation $g(x,y) = c$. Then to solve the constrained optimization problem
find the points $(x,y)$ that solve the equation $\nabla f(x,y) = \lambda \nabla g(x,y)$ for some constant $\lambda$ (the number $\lambda$ is called the Lagrange multiplier). If there is a constrained maximum or minimum, then it must be such a point.