Value of partial derivative depends on all inputs

Statement

The general expression for the partial derivative of a function with respect to any of the inputs is an expression in terms of all the inputs. For instance, the general expression for $f_{x}(x,y)$ is an expression involving both $x$ and $y$. This is because, even though the $y$-coordinate is kept constant when calculating the partial derivative at a particular point, that constant value need not be the same at all the points.

Example

For instance, consider:

$f(x,y):=x^{2}+y^{2}+xy^{2}$

Then, we have:

$f_{x}(x,y)=2x+y^{2}$

and:

$f_{y}(x,y)=2y+2xy$

Note that each of the expressions involves both the variables $x$ and $y$. In particular, this means that the value of $f_{x}$ at a point depends on both the $x$-coordinate and the $y$-coordinate of the point. Thus, for instance, $f_{x}(2,3)=13$ and $f_{x}(2,4)=20$ are distinct because of the different $y$-values.

Related ideas

Second-order mixed partial

The second-order mixed partial derivative captures precisely this fact. Basically, the second-order mixed partial derivative with respect to two of the input variables describes how the partial derivative with respect to one variable changes in terms of the second variable. The statement here can thus be interpreted as saying that the second-order mixed partial derivative of a function is not always zero.

Conditions where the value depends only on the specific input

The only cases where the partial derivative with respect to one variable depends only on that variable is where the function is additively separable in terms of a function purely of that variable and a function of the other variables. Another way of thinking of this is that the second-order mixed partials with respect to that particular variable and all the other variables are zero.

For instance, consider a function of three variables:

$f(x_{1},x_{2},x_{3}):=x_{2}^{2}-x_{1}^{2}x_{3}$

This function is a sum of a function purely of $x_{2}$ and a function that does not involve $x_{2}$. The partial derivative with respect to $x_{2}$ thus involves only $x_{2}$:

$f_{x_{2}}(x_{1},x_{2},x_{3})=2x_{2}$

On the other hand, because $f$ does not have this form with respect to $x_{1}$, the first partial derivative with respect to $x_{1}$ does depend on other variables. Specifically, it depends on both $x_{1}$ and $x_{3}$ in this case:

$f_{x_{1}}(x_{1},x_{2},x_{3})=-2x_{1}x_{3}$