Partial Derivatives

Definition of a Partial Derivative

Let f(x,y) be a function of two variables.  Then we define the partial derivatives  as
 Definition of the Partial Derivative if these limits exist.

Algebraically, we can think of the partial derivative of a function with respect to x as the derivative of the function with y held constant.  Geometrically, the derivative with respect to x at a point P represents the slope of the curve that passes through P whose projection onto the xy plane is a horizontal line.  (If you travel due East, how steep are you climbing?)

Example

Let

f(x,y) = 2x + 3y

then We also use the notation fx  and fy for the partial derivatives with respect to x and y respectively.

Exercise:

Find fy for the function from the example above.

Finding Partial Derivatives the Easy Way

Since a partial derivative with respect to x is a derivative with the rest of the variables held constant, we can find the partial derivative by taking the regular derivative considering the rest of the variables as constants.

Example

Let

f(x,y)  =  3xy2 - 2x2y

then

fx  =  3y2 - 4xy

and

fy  =  6xy - 2x2

Exercises

Find both partial derivatives for

1. f(x,y) = xy sin x

2.                 x + y
f(x,y) =
x - y

Higher Order Partials

Just as with function of one variable, we can define second derivatives for functions of two variables.  For functions of two variables, we have four types:

fxx,     fxy     fyx     and     fyy

Example

Let

f(x,y)  =  y ex

then

fx  =  yex

and

fy  =  ex

Now taking the partials of each of these we get:

fxx = y ex        fxy = ex        fyx = ex       and       fyy = 0

Notice that

fxy  =   fyx

 Theorem   Let f(x,y) be a function with continuous second order derivatives, then                fxy  =   fyx

Functions of More Than Two Variables

Suppose that

f(x,y,z)  =  xy - 2yz

is a function of three variables, then we can define the partial derivatives in much the same way as we defined the partial derivatives for three variables.

We have

fx = y         fy = x - 2z      and       fz = -2y

Application:     The Heat Equation

Suppose that a building has a door open during a snowy day.  It can be shown that the equation

Ht  =  c2Hxx

models this situation where H is the heat of the room at the point x feet away from the door at time t.  Show that

H = e-t cos(x/c)

satisfies this differential equation.

Solution

We have

Ht  =  -e-t cos(x/c)

Hx  =  -1/c e-t sin(x/c)

Hxx  =  -1/c2 e-t cos(x/c)

So that

c2Hxx  =  -e-t cos(x/c)

And the result follows.

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