The partial derivative of z = f( x, y ) with respect to x:
The partial derivative of z = f( x, y ) with respect to y:
The partial derivative of a function with respect to x is the rate of change of the function in the x-direction only. We consider y a constant and find the derivative with only x as a variable. For example, given f( x, y ) = x2 + 3xy + y3, fx = 2x + 3y and fy = 3x + 3y2, which represent the rates of change of the function in the x and y directions, respectively. Of course, second partials may be calculated in like manner yielding fxx = 2, fxy = 3, fyx = 3, fyy = 6y. The subscripts are evaluated from left to right and, as long as the function and its partials are continuous, fxy = fyx.
Also note that fxy = 
. Given a function w = f( x, y, z
), the partials are defined in the same manner with each partial
representing the change in the function in the x, y and z directions,
respectively.
Differentials
If z = f( x, y ), then
, where dz is the total differential of z. If
function z is a function of x and y and f and its first partial
derivatives are continuous, then f is differentiable. The total
differential will be used to define partial derivatives in terms of
parameters and may be utilized to approximate the change in z. In
Calculus I, you could approximate the change in y by the
approximation 
. This is the
change of the function along the tangent line. For functions of 2 or
3 variables we can use the approximations:
to approximate the change in z = f( x, y ) and w = f( x, y, z ).
Propagated Error
This is how propagated error may be calculated when measured error is used to evaluate a formula. For example, if the volume of a right circular cylinder, V = pr2h, is calculated from a measured radius, r, with error of 10%, and height, h , with error of 5%, then the propagated error is approximately:
The relative propagated error is 25%.
Chain rules
Let z = f( x, y ), x = g( t ) and y = h( t ), then
. This is analagous to the
derivative form of Calculus I whereas the total differential is the
differential form. These form may be extended to a function w = f( x,
y, z ) and any number of variables. Of course we could find z in
terms of t by substitution and calculate the derivative
directly.
If z = f( x, y ), x = g( s, t ) and y = h( s, t ),
then 
.
Implicit Differentiation
The chain rules may be applied to finding
derivatives implicitly. Recall from Calculus I that to find the
derivative by implicit differentiation we had to take derivatives
with respect to x and solve for dy/dx. We could do the same here, but
there is a trick that can be useful. If F( x, y ) = 0 defines a
function of two variables, then 
. If
F( x, y, z ) = 0 defines a function of three variables then
For example, if x2 + y2 + z2 = 1, then F( x, y, z ) =
x2 +
y2 +
z2 - 1 and
= -x/z. You could have solved for z
in terms of x and y and found the partial or you could have done a
traditional implicit differentiation by taking the partial
derivatives with respect to x ( remember to treat y as a constant
with x and z variables ) and solving for the partial of z with
respect to x.
