Part 1: Partial Derivatives

Just as derivatives can be used to explore the properties of functions of 1 variable, so also derivatives can be used to explore functions of 2 variables. In this section, we begin that exploration by introducing the concept of a partial derivative of a function of 2 variables.

In particular, we define the partial derivative of f(x,y) with respect to x to be

fx( x,y) = lim h® 0   f( x+h,y)-f( x,y) h

when the limit exists. That is, we compute the derivative of f(x,y) as if x is the variable and all other variables are held constant. To facilitate the computation of partial derivatives, we define the operator

¶ ¶x = ``The partial derivative with respect to x¢¢

Alternatively, other notations for the partial derivative of f with respect to x are

fx( x,y) =  ¶f ¶x =  ¶ ¶x f( x,y) = ¶x f( x,y)

EXAMPLE 1    Evaluate f_x when f( x,y) = x²y+y².       

Solution: To do so, we write

fx( x,y) =  ¶ ¶x (x2y+y2) =  ¶ ¶x x2y+  ¶ ¶x y2

and then we evaluate the derivative as if y is a constant. In particular,

fx( x,y) = y  ¶ ¶x x2+  ¶ ¶x y2 = y·2x+0

That is, y factors to the front since it is considered constant with respect to x. Likewise, y² is considered constant with respect to x, so that its derivative with respect to x is 0. Thus, f_x(x,y) = 2xy.

Likewise, the partial derivative of f( x,y) with respect to y is defined

fy( x,y) = lim h® 0   f( x,y+h)-f( x,y) h

when the limit exists. That is, we evaluate f_y as if y is varying and all other quantities are constant. Moreover, we also define the operator

¶ ¶y = ¢¢The partial derivative with respect to y¢¢

and we often use other notations for f_y( x,y) :

fy( x,y) =  ¶f ¶y =  ¶ ¶y f( x,y) = ¶y f( x,y)

EXAMPLE 2    Find f_x and f_y when f( x,y) = ysin( xy)       

Solution: To find f_x, we use the chain rule

fx = y  ¶ ¶x sin( xy) = ycos(xy)  ¶ ¶x xy = y2cos( xy)

However, to find f_y, we begin with the product rule:

fy =  ¶ ¶y [ ysin( xy) ] = æ è  ¶ ¶y y ö ø sin( xy) +y  ¶ ¶y sin( xy)

We then use the chain rule to evaluate ¶_ysin( xy) :

fy = sin( xy) +ycos( xy)  ¶ ¶y ( xy) = sin( xy) +xycos( xy)

The partial derivatives are equal to the slopes of the two tangent lines, respectively.

EXAMPLE 3    Find f_x and f_y when

f( x,y) = tan-1 æ è  y x ö ø

Solution: To find f_x, we begin with the chain rule,

fx =  ¶ ¶x tan-1( input) =  1 ( input) 2+1     ¶ ¶x ( input)

where the input is y/x. Writing the input as x^-1y and substituting then yields

fx =  1 ( x-1y) 2+1     ¶ ¶x ( x-1y) =  -x-2y x-2y2+1

To simplify this expression, we multiply the numerator and the denominator by x²:

fx =   x2( -x-2y) x2( x-2y2+1) =  -x2x-2y x2x-2y2+x2 =  -y y2+x2

            To find f_y, we again begin with the chain rule,

fy =   ¶ ¶y tan-1( input) =  1 ( input) 2+1     ¶ ¶y ( input)

where the input is y/x. The result is that

fy =    1 æ è  y x ö ø 2   +1     ¶ ¶y æ è  y x ö ø   =    1 æ è  y x ö ø 2   +1 é ë     1 x  ¶ ¶y ( y) ù û

which simplifies to

fy =    1 æ è  y2 x2 +1 ö ø x   =    x æ è  y2 x2 +1 ö ø x2   =    x x2+y2