directional derivative
Let
and let
a
,
v
∈
R
n
, then the directional derivative of
f
at
a
along
v
is defined as
∇
v
f
(
a
)
:=
lim
h
→
0
f
(
a
+
h
v
)
−
f
(
a
)
h
Frozen Function
Suppose that
f
:
R
n
→
R
and
a
∈
R
n
then we can define a new function
f
i
,
a
:
R
→
R
by
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
∂
f
∂
x
i
(
a
)
=
d
d
h
[
f
i
,
a
(
h
)
]
TODO
Todo
Consider the function
f
(
x
,
y
,
z
)
=
x
2
+
y
2
+
z
2
, and the point
(
a
1
,
a
2
,
a
3
)
∈
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
∂
∂
y
f
(
x
,
y
,
z
)
=
2
y
Gradient
The gradient of a function
f
:
R
n
→
R
at the point
p
is defined as
∇
f
(
p
)
:=
(
∂
f
∂
(
x
1
)
(
p
)
,
∂
f
∂
(
x
2
)
(
p
)
,
…
,
∂
f
∂
(
x
n
)
(
p
)
)
Todo
Suppose we have the fnction
f
(
x
,
y
)
=
sin
(
x
)
, and
(
p
1
,
p
2
)
∈
R
2
show that
∇
f
(
p
)
=
(
cos
(
p
1
)
,
0
)
By definition
∇
f
(
p
)
=
(
∂
f
∂
x
1
(
p
)
,
∂
f
∂
x
2
(
p
)
)
. Since we know that
∂
f
∂
x
1
=
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
)
so
p
=
(
p
1
,
p
2
)
so
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
∂
f
∂
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
∇
f
(
p
)
=
(
cos
(
p
1
)
,
0
)
differentaibility
a function
f
:
A
→
R
m
with
A
⊆
R
n
is differentiable at the point
a
∈
R
n
if there exists a linear transformation
T
:
R
n
→
R
m
such that
lim
x
→
a
‖
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
ToDo