When you find the partial derivative of a function with respect to a particular variable (say, #x#), then you treat every other variable like it's a constant. You then perform the derivative exactly as before. It takes a few problems before it "clicks", but the process will soon feel exactly the same as regular differentiation.

We have #f(x, y, z, t) = (xy)/(t + 2z)#.

We'll first find #(delf)/(delx)#, which can be more conveniently notated #f_x#. Both notations refer to the first partial derivative of #f# with respect to #x#. For #f_x#, we treat #x# like a variable and everything else like a regular number. Thus, #f = (y/(t+2z))(x)# and the leftmost term is considered constant. Because the derivative of the function #Cx# is #C#, where #C# is constant, it follows that #f_x = y / (t + 2z)#.

A similar procedure is followed to find #f_y#. Since #f = (x/(t + 2z))(y)#, it follows that #f_y = x / (t+2z)#.

#f_z# is found by seeing that #f = (xy)(t + 2z)^(-1)#. (Remember, #x, y, t# are constant.) Thus #f_z = (-1)(xy)(t + 2z)^(-2)(2)#. Note that we used the chain rule. This can be simplified as #f_z = (-2xy)/(t + 2z)^2#.

Just as we found #f_z#, #f_t = (-1)(xy)(t + 2z)^(-2)#. This can be written #f_t = (-xy)/(t+2z)^2#.

If you still find the process difficult, simply rewriting the equation with capital letters except in the variable you're differentiating with respect to, can help. For example #f_z# is more easily calculated if you write #f = (XY)/(T + 2z)#. Mentally, it is more clear that #x, y, t# are constants.