f(x,y) = x2y y3 2y

1
Definition of partial derivative (page 128-129)
f(x,y) x2y y3 2y
f fx x
f fy y
2xy
x2 3y2 2y ln2
z f(x,y) cos(xy) x cos2y 3
z (x,y) x
y sin(xy) cos2y
x sin xy 2x cos y sin y
fy(x,y)
z (x0,y0) x
y0 sin(x0y0) cos2y0
fy(2,p/2)
2 sin p 4 cos (p/2) sin (?/2)
0
y 3(xy)2/3
x 3(xy)2/3
z f(x,y) (xy)1/3 fx fy
2
Compare the functions y x2 and y x2/3 in R2,
and decide whether or not each function is
differentiable at x 0.
y
y
Either by looking at graphs or from the
definition of derivative, we see that y x2 is
differentiable at x 0 but y x2/3 is not.
y x2
y x2/3
x
x
Now consider the function z f(x,y) (xy)1/3 in
R3.
What does f(x,y) look like in the xz plane where
y 0?
z 0
z
What does f(x,y) look like in the yz plane where
x 0?
z 0
y
What does f(x,y) look like in the plane where x
y?
z t2/3 where t represents a variable which
gives the value of x y.
x
3
Now consider the function z f(x,y) (xy)1/3 in
R3.
What does f(x,y) look like in the xz plane where
y 0?
z 0
z
What does f(x,y) look like in the yz plane where
x 0?
z 0
y
What does f(x,y) look like in the plane where x
y?
z t2/3 where t represents a variable which
gives the value of x y.
x
We find that for functions in R3, it is possible
for a level curve in one direction to be
differentiable at a point while a level curve
in another direction is not differentiable at the
same point.
A function in R2 is differentiable at a point, if
and only if there exists a line tangent to the
function at the point.
A function in R3 is differentiable at a point, if
and only if there exists a plane tangent to the
function at the point.
4
In order that a function f(x,y) be differentiable
at a point (x0 , y0), the function must be
smooth in all directions, not just in the
x-direction and y-direction.
If a function f(x,y) is differentiable at point
(x0 , y0), then we can use the value of each
partial derivative at the point to find the
equation of the plane tangent to the function at
the point
The equation of the tangent plane can be written
as g(x,y) ax by c.
The following must be true
g(x0 , y0) ax0 by0 c f(x0 , y0) ,
gx(x0 , y0) a fx(x0 , y0) ,
gy(x0 , y0) b fy(x0 , y0) .
The equation of the tangent plane is
g(x,y) fx(x0 , y0) x fy(x0 , y0) y f(x0 ,
y0) fx(x0 , y0) x0 fy(x0 , y0) y0 which can
be written
z f(x0 , y0) fx(x0 , y0) (x x0) fy(x0 ,
y0) (y y0) .
5
Definition of differentiable and tangent plane in
R3 (page 133) See the derivation on page 132.
Example Find the plane tangent to the graph of z
x2 y3 cos(?xy) at the point (8 , 4 , 1) .
f(8 , 4)
1
f fx x
f fy y
2x ?y sin(?xy)
3y2 ?x sin(?xy)
fx(8 , 4)
fy(8 , 4)
16
48
The equation of the tangent plane is
z 1 16(x 8) 48(y 4) which can be
written
16x 48y z 63 .
6
Look again at the definition of differentiable in
R3 on page 133, and observe that we can say
f(x,y) is differentiable at (x0 , y0) if
This is the difference between the exact function
value at point (x , y) and the approximated value
from the plane tangent to the function at (x0 ,
y0).
(x x0) (y y0)
f(x,y) f(x0 , y0) fx(x0 , y0) fy(x0 , y0)
lim 0
(x , y)?(xo , y0)
(x , y) (xo , y0)
If x (x1 , x2 , , xn) is a vector in Rn, and
f(x) is a function from Rn to R, then we define
f(x) to differentiable at x0 if
f(x) f(x0) Df(x0) (x x0)
lim 0 where
x?xo
x x0
Df(x0) is the 1?n matrix of partial derivatives
at x0 , and (x x0) is the n?1 matrix consisting
of the differences between each variable and its
specific value at x0 . (NOTE A 1?n matrix times
an n?1 matrix is the same as the dot product of
two vectors.
7
Suppose x (x1 , x2 , , xn) is a vector in Rn,
and f(x) is a function from Rn to Rm. Then, we
can write f(x) f1(x) , f2(x) , , fm(x) .
We let Df(x) represent the m?n matrix with row i
consisting of the following partial derivatives
?fi ?fi ?fi . ?x1 ?x2
?xn
We call Df(x) the derivative (matrix) of f , and
of course Df(x0) is the derivative (matrix) of f
at x0 .
Look at the general definition of differentiable
on page 134.
8
Example Find the derivative matrix for w
f(x,y,z) ( x2 xy4eyz , ln(xzy) ) .
_ _
Df(x,y,z)

_ _
2x y4eyz
4xy3eyz xy4zeyz
xy5eyz
z / (xzy)
1 / (xzy)
x / (xzy)
Find the linear approximation at the point x0
(1 , e , 0).
x 1 y e z 0
w f (x0) Df (x0) (x x0)
f (1 , e , 0) Df (1 , e , 0)
9
2 e4 4e3 e5 0 1 / e 1 / e
x 1 y e z 0
1 e4 1
1 e4 1
(2 e4)x 2 e4 4e3y 4e4 e5z y / e 1
z / e


(2 e4)x 4e3y e5z 1 4e4 (y z) / e
Observe that since w f(x,y,z) goes from R3 to
R2, this linear approximation is really two
linear approximations, one for each of the
component functions. That is,
10
Observe that since w f(x,y,z) goes from R3 to
R2, this linear approximation is really two
linear approximations, one for each of the
component functions. That is,
The linear approximation for f1(x,y,z) x2
xy4eyz is
w1 (2 e4)x 4e3y e5z (1 4e4) .
The linear approximation for f2(x,y,z) ln(xzy)
is
w2 (1/e)y (1/e)z .
11
Recall that for a function f from Rn to R1, the
derivative matrix is 1?n. In certain situations,
it will be convenient to treat this 1?n matrix as
a vector. See the definition of a gradient on
page 136.
Look at Theorem 8 on page 137. Look at Theorem 9
on page 137. Look at the chart at the top of page
138.