Defn: Let f : R^{n} → R^{m}. Define g_{ij} : R^{1} → R^{1}, g_{ij}(t) = f_{i}(x_{1}, ..., x_{j−1}, t, x_{j+1}, ..., x_{n}). If g is differentiable at a, then the partial derivative of f_{i} is defined by

\frac{∂f_{i}}{∂x_{j}} (a) = lim_{h→0} \frac{f_{i}(a_{1},...,a_{j−1},a_{j}+h,a_{j+1},...,a_{n})−f_{i}(a_{1},...,a_{j−1},a_{j},a_{j+1},...,a_{n})}{h}

= lim_{h→0} \frac{f_{i}(a+he_{j})−f_{i}(a)}{h}

Ex: f : R^{2} → R, f(x, y) = \{ \begin{matrix}0 & (x, y) = (0, 0) \\ \frac{xy}{x^{2}+y^{2}} & otherwise \end{matrix}

\frac{∂f}{∂x}(0, 0) = lim_{h→0} \frac{f(0+h,0)−f(0,0)}{h} = lim_{h→0} \frac{0−0}{h} = 0

\frac{∂f}{∂y} (0, 0) = lim_{h→0} \frac{f(0,0+h)−f(0,0)}{h} = lim_{h→0} \frac{0−0}{h} = 0

BUT f is not continuous at (0, 0)!!!!!!!!!

Defn: Suppose f : R^{1} → R^{1}. The derivative of f at the point a is f′(a) = lim_{h→0} \frac{f(a+h)−f(a)}{h} if above limit exists.

f′(a) is the derivative of f at x if

lim_{h→0} \frac{f(a + h) − f(a) − f′(a)h}{h} = 0

lim_{x→a} \frac{f(x) − f(a) − f′(a)(x − a)}{(x − a)} = 0

y = f′(a)[x − a] + f(a) is the linear approximation of f near a.

Defn: Let V and W be vector spaces. A linear transformation from V to W is a function T : V → W that satisfies the following two conditions. For each u and v in V and scalar a, i.) T (au) = aT (u) ii.) T (u + v) = T (u) + T (v) Thm: Let T : V → W be a linear transformation. Then T (0) = 0 Pf: T (0) = T (0 + 0) = T (0) + T (0) Thm: Let A be an m × n matrix. Then the function

\begin{matrix}T : R^{n} → R^{m} \\ T (x) = Ax \end{matrix}

is a linear transformation. Thm: If T : R^{n} → R^{m} is a linear transformation, then T (x) = Ax where

A = [T (e_{1})...T (e_{n})]

Ex: If T : R^{2} → R and T (1, 0) = 3, T (0, 1) = 4, then

T (x, y) = xT (1, 0) + yT (0, 1) = (3, 4) ( \begin{matrix}x \\ y \end{matrix} ) = 3x + 4y

Defn: Suppose f : R^{1} → R^{1} is differentiable at a. Then

lim_{h→0} \frac{f(a + h) − f(a) − f′(a)h}{h} = 0

lim_{x→a} \frac{f(x) − f(a) − f′(a)(x − a)}{(x − a)} = 0

Defn: Suppose A ⊂ R^{n}, f : A → R^{m}.

f is said to be differentiable at a point a if there exists an open ball V such that a ∈ V ⊂ A and a linear function T such that

lim_{h→0} \frac{||f(a + h) − f(a) − T (h)||}{||h||} = 0

lim_{x→a} \frac{||f(x) − f(a) − T (x − a)||}{||x − a||} = 0

Suppose f : R^{n} → R.

Then T = (b_{1}, ..., b_{n}) and T x = (b_{1}, ..., b_{n}) ( \begin{matrix}x_{1} \\ x_{1} \\ ... \\ x_{n} \end{matrix} ) = Σb_{i}x_{i}

lim_{x→a} \frac{||f(x) − f(a) − Σb_{i}(x_{i} − a_{i})||}{||x − a||} = 0

f : R → R:

lim_{x→a} \frac{f(x) − f(a) − f′(a)(x − a)}{(x − a)} = 0

y = f′(a)[x − a] + f(a) is the linear approximation of f near a.

\frac{df}{dx}(a) = f′(a)

f : R^{n} → R:

lim_{x→a} \frac{f(x) − f(a) − Σb_{i}(x_{i} − a_{i})}{||x − a||} = 0

y = f(a) + Σb_{i}(x_{i} − a_{i}) approximates y = f(x)

f(x) = f(a) + Σb_{i}(x_{i} − a_{i}) + ||x − a||r(x, a)

where lim_{x→a}r(x, a) = 0

(df)_{a} = Σb_{i}(x_{i} − a_{i})

Mean Value Theorem: Suppose 1.) f continuous on [a, b] 2.) f differentiable on (a, b)

Then there exists c ∈ (a, b) such that f′(c) = \frac{f(b)−f(a)}{b−a}

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