derivative

let $\text{[math]}$ be an open subset of $\text{[math]}$ , and let $\text{[math]}$ be a function. then $\text{[math]}$ is differentiable at $\text{[math]}$ with derivative $\text{[math]}$ if the limit
$\text{[math]}$ [cite:@calc_hubbard_2015]

another derivative notation $\text{[math]}$ to denote the $\text{[math]}$ th derivative of a function we write $\text{[math]}$ , this reads "the $\text{[math]}$ th derivative of $\text{[math]}$ with respect to $\text{[math]}$ "

an alternative definition of the derivative in one dimension
let $\text{[math]}$ be an open subset of $\text{[math]}$ and $\text{[math]}$ a function. then $\text{[math]}$ is differentiable at $\text{[math]}$ , with derivative $\text{[math]}$ , if and only if
$\text{[math]}$ the letter $\text{[math]}$ denotes the "change in"; $\text{[math]}$ is the change in the function; $\text{[math]}$ is the change in the variable $\text{[math]}$ . the function $\text{[math]}$ that multiplies $\text{[math]}$ by the derivative $\text{[math]}$ is thus a linear function of the change in $\text{[math]}$ .
[cite:@calc_hubbard_2015 definition 1.7.5]

let $\text{[math]}$ be an open subset and let $\text{[math]}$ be a mapping; let $\text{[math]}$ be a point in $\text{[math]}$ . if there exists a linear transformation $\text{[math]}$ such that
$\text{[math]}$ then $\text{[math]}$ is differentiable at $\text{[math]}$ , and $\text{[math]}$ is unique and is the derivative of $\text{[math]}$ at $\text{[math]}$ , denoted $\text{[math]}$ , and whose transformation matrix is of dimensions $\text{[math]}$ .
[cite:@calc_hubbard_2015 proposition and definition 1.7.9 (derivative)]

if $\text{[math]}$ is differentiable at $\text{[math]}$ , then all partial derivatives of $\text{[math]}$ exist, and the matrix representing $\text{[math]}$ is $\text{[math]}$ (the jacobian matrix).

since the square of an $\text{[math]}$ matrix is another $\text{[math]}$ matrix, and such a matrix can be "identified" with $\text{[math]}$ , this could be written as a function $\text{[math]}$ . this is one time when a linear transformation is easier to deal with than the corresponding matrix. we denote by $\text{[math]}$ the set of $\text{[math]}$ matrices, and consider the squaring map
$\text{[math]}$ in this case we can compute the derivative without computing the jacobian matrix. we shall see that $\text{[math]}$ is differentiable and that its derivative $\text{[math]}$ is the linear transformation that maps $\text{[math]}$ to $\text{[math]}$ :
$\text{[math]}$ the first thing to realize is that the map
$\text{[math]}$ is a linear transformation. the asseration is that
$\text{[math]}$ since $\text{[math]}$ ,
$\text{[math]}$ this gives
$\text{[math]}$ so $\text{[math]}$ is indeed the derivative.
[cite:@calc_hubbard_2015 example 1.7.17]

vector calculus derivative table

let $\text{[math]}$ be open.

if $\text{[math]}$ is a constant function, then $\text{[math]}$ is differentiable, and its derivative is $\text{[math]}$ (the zero linear transformation $\text{[math]}$ , represented by the $\text{[math]}$ matrix filled with 0's)
if $\text{[math]}$ is linear, then it is differentiable everywhere, and its derivative at all points $\text{[math]}$ is $\text{[math]}$ , i.e., $\text{[math]}$ .
if $\text{[math]}$ are $\text{[math]}$ scalar-valued functions differentiable at $\text{[math]}$ , then the vector-valued mapping $\text{[math]}$ is differentiable at $\text{[math]}$ , with derivative $\text{[math]}$ conversely, if $\text{[math]}$ is differentiable at $\text{[math]}$ , each $\text{[math]}$ is differentiable at $\text{[math]}$ , and $\text{[math]}$ .
if $\text{[math]}$ are differentiable at $\text{[math]}$ , then so is $\text{[math]}$ , and $\text{[math]}$ 5. if $\text{[math]}$ and $\text{[math]}$ are differentiable at $\text{[math]}$ , then so is $\text{[math]}$ , and the derivative is given by $\text{[math]}$ 6. if $\text{[math]}$ and $\text{[math]}$ are differentiable at $\text{[math]}$ , and $\text{[math]}$ , then so is $\text{[math]}$ , and the derivative is given by $\text{[math]}$ 7. if $\text{[math]}$ are both differentiable at $\text{[math]}$ , then so is the dot product $\text{[math]}$ , and (as in one dimension) $\text{[math]}$

[cite:@calc_hubbard_2015 theorem 1.8.1]