Download 7.4 Partial Derivatives

7.4 Partial Derivatives
Let’s focus on functions z = f (x, y), i.e. functions of two variables,
but everything we say can be generalized to functions of 3 or more
variables. Define the partial derivative of f with respect to x as
∂f
f (x0 + h, y0 ) − f (x0 , y0 )
(x0 , y0 ) = lim
h→0
∂x
h
So what is happening here is we are fixing y0 and letting x vary in
the sense that x0 + h → x0 . This gives us the rate of change of
the function f (x, y) in the x−direction. Similarly we can define
the partial derivative of f with respect to y as
∂f
f (x0 , y0 + h) − f (x0 , y0 )
(x0 , y0 ) = lim
h→0
∂y
h
Sample partial derivatives
Let f (x, y) = x3 y 4 then
∂f
(x + h)3 y 4 − x3 y 4
(x + h)3 − x3 4
(x, y) = lim
= lim
y
h→0
h→0
∂x
h
h
d 3 4
x y
=
dx
2 4
Therefore it is easy to see that ∂f
∂x (x, y) = 3x y . This simple
example illustrates a more general rule which is that when you are
taking the partial derivative with respect to x, you treat y as fixed,
or as a constant. So what would
∂f
(x, y)?
∂y
∂f
(x, y) = 4x3 y 3
∂y
Practice some examples
We will go over the answers to these in class. For each function
∂f
f (x, y) find ∂f
dx and ∂y : Don’t forget the chain rule, product rule,
and quotient rule.
f (x, y) = xn y m (here m, n are any real numbers)
f (x, y) = xye1−x
2 +y 2
f (x, y) = x ln(2x − y 2 )
f (x, y) =
2x2
x2 + y 2
Examples, solutions
f (x, y) = xn y m (here m, n are any real numbers)
∂f
∂f
= nxn−1 y m and
= mxn y m−1
∂x
∂y
f (x, y) = xye1−x
2 +y 2
∂f
2
2
2
2
= ye1−x +y − 2x2 ye1−x +y
∂x
∂f
2
2
2
2
and
= xe1−x +y + 2xy 2 e1−x +y
∂y
Examples, solutions
f (x, y) = x ln(2x − y 2 )
2x
∂f
= ln(2x − y 2 ) +
∂x
2x − y 2
f (x, y) =
and
∂f
−2xy
=
∂y
2x − y 2
2x2
x2 + y 2
∂f
4x(x2 + y 2 ) − 4x3
=
∂x
(x2 + y 2 )2
and
∂f
−4x2 y
= 2
∂y
(x + y 2 )2
Interpreting partial derivatives as slope
As in single variable calculus, the partial derivative represents the
slope, though now it is the slope of a tangent plane at a point on a
surface, and there are two slopes, representing the change in the
∂f
x−direction ( ∂f
∂x ) and the change in the y−direction ( ∂y ). For any
function z = f (x, y) you can ask for the values of the slope of the
tangent plane at a point, so given a point (x0 , y0 ) you will
∂f
compute ∂f
∂x (x0 , y0 ) and ∂y (x0 , y0 ). For example:pFind the slopes
of the tangent plane to the surface of f (x, y) = 1 − x2 − y 2 at
(x, y) = (0, 0). Interpret this graphically. We will go over this in
class. What if the point in question was (x, y) = (1/2, 1/2)? or
(x, y) = ( √12 , √12 )?