Computation

To compute a partial derivative, we simply treat all variables in the function as being constant, except for the variable which appears in the partial derivative. Thus, to compute $\frac{\partial f}{\partial x}$ for f(x,y,z) we treat y and z as constants. Here's a quick example. Let g(x,y) = x² + y².

$$\frac{\partial g}{\partial x} = \frac{\partial }{\partial x}... ...^2}{\partial x} + \frac{\partial y^2}{\partial x} = 2x + 0 = 2x$$

As you can see, if we treat all of the variables except one as a constant, then we are really only differentiating a function of one variable. It should not surprise you then that all of the familiar rules of differentiation from one variable calculus apply here. Thus, the product, power, and quotient rule still remain valid, as do all of the other rules and tricks for differentiation.

Again, to emphasize the point, when we compute $\frac{\partial f}{\partial y}$ we treat x as a fixed constant. Let h(x,y) = xy + 2y² + x²e^y.

We can evaluate these derivatives at a single point or leave them as functions. Thus,

$$\left.\frac{\partial h}{\partial y}\right\vert _{(1,-1)} = \... ...^y)\right\vert _{(1,-1)} = 1 + 4(-1) + 1^2 e^{-1} = e^{-1} - 3.$$

We can carry out the same computations with functions of any number of variables. If $z = f(x_1, x_2, \dots, x_n)$ then $\frac{\partial f}{\partial x_i}$ is the slope of the function in the direction of positive x_i axis. For example, if we let w = x² + 2y² - 5z² then

$$\frac{\partial w}{\partial x} = 2x \qquad \frac{\partial w}{\partial y} = 4y \qquad \frac{\partial w}{\partial z} = -10z.$$