Let f : R^{n} → R. Define g_{j} : R → R by g_{i}(t) = f_{i}(x_{1}, ..., x_{j−1}, t, x_{j+1}, ..., x_{n}). If g is differentiable at a_{j}, then the partial derivative of f with respect to x_{j} is defined by \frac{∂f_{i}}{∂x_{j}} (x) = g′(x_{j}) = lim_{h→0} \frac{g(x_{j}+h)−g(x_{j})}{h} = lim_{h→0} \frac{f_{i}(x_{1},...,x_{j−1},x_{j}+h,x_{j+1},...,x_{n})−f_{i}(x_{1},...,x_{j−1},x_{j},x_{j+1},...,x_{n})}{h}

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

Ex: \frac{∂(x^{2}y)}{∂x} = \frac{∂(x^{2}y)}{∂x} =

The gradient of f is denoted by ∇f(a) = (\frac{∂f_{1}}{∂x_{1}} (a), ..., \frac{∂f_{1}}{∂x_{n}} (a))

Ex: ∇x^{2}y = When is f differentiable (not just partially differentiable)?

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)!!!!!!!!! RECALL: f differentiable implies f continuous

Ex: g : R^{2} → R, g(x, y) = x + 2y \frac{∂g}{∂x}(0, 0) = lim_{h→0} \frac{f(0+h,0)−f(0,0)}{h} = lim_{h→0} \frac{h−0}{h} = 1 \frac{∂g}{∂y} (0, 0) = lim_{h→0} \frac{f(0,0+h)−f(0,0)}{h} = lim_{h→0} \frac{2h−0}{h} = 2 We will see later that g is differentiable.

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

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

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

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

or equivalently,

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

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

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

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

Thus when x is close to a, then f(x) is close to f(a)+f′(a)[x−a] Thus y = f(a) + f′(a)[x − a] is the best linear approximation of f near x = 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

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

Thus when x is close to a, then f(x) is close to f(a)+f′(a)[x−a]

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

Hence the tangent line to f at a is y = f(a) + f′(a)[x − a]

The tangent line can be written as a constant plus a linear function.

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_{2} ... b_{n}) and T x = (b_{1} b_{2} ... b_{n}) ( \begin{matrix}x_{1} \\ x_{2} \\ . \\ . \\ . \\ x_{n} \end{matrix} ) = Σb_{i}x_{i}

Also, T (x − a) = (b_{1} b_{2}... b_{n}) ( \begin{matrix}x_{1} − a_{1} \\ x_{2} − a_{2} \\ . \\ . \\ . \\ x_{n} − a_{n} \end{matrix} ) = Σb_{i}(x_{i} − a_{i})

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

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

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)