DEFINITION:

The partial derivative of f (x, y) with respect to x at the point (x ₀, y₀) is

provided the limit exists.

FACT:

The slope of the curve z = f (x, y ₀) at the point P (x ₀, y ₀, f (x ₀, y ₀)) in the plane y = y ₀ is the value of the partial derivative of f with respect to x at (x ₀, y ₀). This slope is the slope of the tangent line to the curve in the plane y = y ₀.

FACT:

The following notations are equal.

How do you find f _x of a function f (x, y)? Well, first you treat the variable y like a constant. So when you take a derivative of y, you will get zero. The derivative of x is just like how we have been doing the past two semesters.

EXAMPLE 1:

Find f _x for f (x, y) = x ² - xy + y ².

SOLUTION:

Before taking the partial derivative of f, notice the -xy term. Treat this term like you would -2x. The derivative of -2x is -2, so the derivative of -yx is -y.

f _x = 2x - y + 0 = 2x - y

EXAMPLE 2:

Find f _x for f (x, y) = (xy - 1) ².

SOLUTION:

Here we will have to use the chain rule.

f _x = 2(xy - 1)(y) = 2y(xy - 1)

EXAMPLE 3:

Find f _x for f (x, y) = cos ² (3x - y ²).

SOLUTION:

f _x = (2cos (3x - y ²))(-sin (3x - y ²))(3) = -6cos (3x - y ²)sin (3x - y ²)

EXAMPLE 4:

SOLUTION:

We will have to use the quotient rule to find this derivative.

DEFINITION:

The partial derivative of f (x, y) with respect to y at the point (x ₀, y₀) is

provided the limit exists.

FACT:

The slope of the curve z = f (x ₀, y) at the point P (x ₀, y ₀, f (x ₀, y ₀)) in the plane x = x ₀ is the value of the partial derivative of f with respect to y at (x ₀, y ₀). This slope is the slope of the tangent line to the curve in the plane x = x ₀.

FACT:

The following notations are equal.

How do you find f _y of a function f (x, y)? Well, first you treat the variable x like a constant. So when you take a derivative of x, you will get zero. The derivative of y is just like how we have been doing the past two semesters.

EXAMPLE 5:

Find f _y for f (x, y) = x ² - xy + y ².

SOLUTION:

Before taking the partial derivative of f, notice the -xy term. Treat this term like you would -2y. The derivative of -2y is -2, so the derivative of -yx is -x.

f _y = 0 - x + 2y = -x + 2y

EXAMPLE 6:

Find f _y for f (x, y) = (xy - 1) ².

SOLUTION:

Here we will have to use the chain rule.

f _y = 2(xy - 1)(x) = 2x(xy - 1)

EXAMPLE 7:

Find f _y for f (x, y) = cos ² (3x - y ²).

SOLUTION:

f _y = (2cos (3x - y ²))(-sin (3x - y ²))(-2y) = 4y cos (3x - y ²)sin (3x - y ²)

EXAMPLE 8:

SOLUTION:

We will have to use the quotient rule to find this derivative.

Now let us find f _x, f _y, and f _z for a function f (x, y, z).

EXAMPLE 9:

Find f _x, f _y, and f _z for f (x, y, z) = xy + yz + xz.

SOLUTION:

When finding f _x, treat y and z as constants.

f _x = y + 0 + z = y + z

When finding f _y, treat x and z as constants.

f _y = x + z + 0 = x + z

When finding f _z, treat x and y as constants.

f _z = 0 + y + x = y + x

EXAMPLE 10:

Find f _x, f _y, and f _z for the following function.

SOLUTION:

Recall that in calculus I, for a function to have a derivative at a point, the function had to be continuous at the point. This is not necessary for partial derivative. In fact, a function f (x, y) can have partial derivatives with respect to both x and y at a point without being continuous there.

EXAMPLE 11:

Find all second order partial derivatives of f (x, y) = sin xy.

SOLUTION:

f _x = y cos xy

f _y = x cos xy

f _{x x} = (-y sin xy)(y) = -y ²sin xy

f _{y y} = (-x sin xy)(x) = -x ²sin xy

f _{x y} = (-y sin xy)(x) = -xy sin xy

f _{y x} = (-x sin xy)(y) = -xy sin xy

Notice that f _{x y} = f _{y x}. Is this a fluke?

EXAMPLE 12:

Find all second order partial derivatives of f (x, y) = x e ^y + y + 1.

SOLUTION:

f _x = e ^y

f _y = x e ^y + 1

f _{x x} = 0

f _{y y} = x e ^y

f _{x y} = e ^y

f _{y x} = e ^y

Again f _{x y} = f _{y x}. This is not a fluke, and here is the theorem that states why this happens.

 EULER'S THM:

If f (x, y) and its partial derivatives f _x, f _y, f _{x y}, and f _{y x} are defined throughout an open region containing a point (a, b) and all are continuous at (a, b), then f _{x y} (a, b) = f _{y x} (a, b).

 EXAMPLE 13:

Given w = ln (2x + 3y), verify w _{x y} = w _{y x}.

SOLUTION:

Therefore, w _{x y} = w _{y x}.

 EXAMPLE 14:

Find the value of ¶ z /¶ x at the point (1, 1, 1) if the equation xy + z ³ x - 2yz = 0 defines z as a function of two independent variables x and y and the partial derivatives exist.

 SOLUTION:

We are going to do implicit differentiation. Every time we take the derivative of z, tack on  ¶ z /¶ x. y in this case is a constant.