differentiation

directional derivative

Let

f : ℝ^{n} \to ℝ

and let

a, v \in ℝ^{n}

, then the directional derivative of

f

a

along

v

is defined as

\nabla_{v} f (a) : = {lim}_{h \to 0} \frac{f (a + h v) - f (a)}{h}

Partial Derivative

Let

f : ℝ^{n} \to R

then the partial derivative with respect to

x_{i}

is the directional derivative of

f

a

along

e_{i}

\frac{\partial}{\partial x_{i}} f (a) : = \nabla_{e_{i}} f (a)

Frozen Function

Suppose that

f : ℝ^{n} \to ℝ

and

a \in ℝ^{n}

then we can define a new function

f_{i, a} : ℝ \to ℝ

f_{i, a} (h) : = f (a + h e_{i})

The frozen function get's it's name because it can be thought of by taking the sample point $a$ and then freezing in all directions except for along the $i$ -th component of the input for $f$

Partial Derivative becomes frozen function derivative

\frac{\partial f}{\partial x_{i}} (a) = \frac{d}{d h} [f_{i, a} (h)]

Todo

Consider the function

f (x, y, z) = x^{2} + y^{2} + z^{2}

, and the point

(a_{1}, a_{2}, a_{3}) \in ℝ^{3}

then show that

f_{y, a} (h)

equals the function

g (x) = a_{1}^{2} + {(x)}^{2} + a_{3}^{2}

Let's start by looking at the definition of

f_{2, a} (h) = f (a + h e_{2})

now recall that

e_{2} = (0, 1, 0)

, so that

f (a + h e_{2}) = f ((a_{1}, a_{2}, a_{3}) + (0, h, 0)) = f (a_{1}, a_{2} + h, a_{3}) = {(a_{1})}^{2} + {(a_{2} + h)}^{2} + {(a_{3})}^{2}

Todo

compute the partial derivative with resepct to y of

f (x, y, z) = x^{2} + y^{2} + z^{2}

Leaving x and z as constants, and evaluating the derivative as

f (x, y, z) = C + y^{2} + C

, using the sum rule and the power rule, the derivative becomes

\frac{\partial}{\partial y} f (x, y, z) = 2 y

Gradient

The gradient of a function

f : ℝ^{n} \to R

at the point

p

is defined as

\nabla f (p) : = (\frac{\partial f}{\partial} (x_{1}) (p), \frac{\partial f}{\partial} (x_{2}) (p), \dots, \frac{\partial f}{\partial} (x_{n}) (p))

Todo

Suppose we have the fnction

f (x, y) = \sin (x)

, and

(p_{1}, p_{2}) \in ℝ^{2}

show that

\nabla f (p) = (\cos (p_{1}), 0)

By definition

\nabla f (p) = (\frac{\partial f}{\partial x_{1}} (p), \frac{\partial f}{\partial x_{2}} (p))

. Since we know that

\frac{\partial f}{\partial x_{1}} = \frac{d}{d h} (f_{1, p} (h))

, and we know that

f_{1, p} (h) = f (p + h e_{1})

since

h e_{1} = (h, 0)

p = (p_{1}, p_{2})

p + h e_{1} = (p_{1} + h, p_{2})

so then

f (p + h e_{1}) = \sin (p_{1} + h)

, evaluating the limit from the directional derivative, it evaluates to

\cos (p_{1})

we will show that $\frac{\partial f}{\partial x_{2}} (p) = 0$ , we know that $f_{2, p} (h) = f (p + h e_{2}) = \cos (p_{1})$ and thus it is a constant funtion, this implies tha directional derivative is 0

Therefore we can se that $\nabla f (p) = (\cos (p_{1}), 0)$

differentaibility

a function

f : A \to ℝ^{m}

with

A \subseteq ℝ^{n}

is differentiable at the point

a \in ℝ^{n}

if there exists a linear transformation

T : ℝ^{n} \to ℝ^{m}

such that

{lim}_{x \to a} \frac{‖ f (x) - (f (a) + T (x - a)) ‖}{‖ x - a ‖} = 0

continuous partials implies differentiability

If all the partial derivatives of a function exist in a neighborhood of a point

x_{0}

and are continuous at the point

x_{0}

, then the function is differentiable at that point

x_{0}

🏗️ ΘρϵηΠατπ🚧 (under construction)

🏗️ $Θ ρ ϵ η Π α τ π$ 🚧 (under construction)