Definition of partial derivative (page 128-129)
f(x,y) = x2y + y3 + 2y
f
— = fx = 2xy
x
f
— = fy = x2 + 3y2 + 2y ln2
y
z = f(x,y) = cos(xy) + x cos2y – 3
z
—(x,y) = – y sin(xy) + cos2y
x
fy(x,y) = – x sin xy – 2x cos y sin y
z
—(x0,y0) = – y0 sin(x0y0) + cos2y0
x
fy(2,p/2) =
z = f(x,y) =
(xy)1/3
y
fx = ———
3(xy)2/3
– 2 sin p – 4 cos (p/2) sin (p/2) = 0
x
fy = ———
3(xy)2/3
Compare the functions y = x2 and y = x2/3 in R2, and decide whether or
not each function is differentiable at x = 0.
Either by looking at graphs or
y = x2 y
y = x2/3 y
from the definition of derivative,
we see that y = x2 is differentiable
x
x at x = 0 but y = x2/3 is not.
Now consider the function
z = f(x,y) = (xy)1/3 in R3.
z
What does f(x,y) look like in
the xz plane where y = 0? z = 0
What does f(x,y) look like in
the yz plane where x = 0? z = 0
y
x
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.
Now consider the function
z = f(x,y) = (xy)1/3 in R3.
z
What does f(x,y) look like in
the xz plane where y = 0? z = 0
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.
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) .
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(pxy) at the point
(8 , –4 , –1) .
f(8 , –4) = –1
f
— = fx = 2x + py sin(pxy)
x
f
— = fy = 3y2 + px sin(pxy)
y
fx(8 , –4) = 16
fy(8 , –4) = 48
The equation of the tangent plane is
z = –1 + 16(x – 8) + 48(y + 4) which can be written 16x + 48y – z = –63 .
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).
f(x,y) – f(x0 , y0) – fx(x0 , y0) fy(x0 , y0)
(x – x0)
(y – 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 – x0 ||
xxo
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.
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.
Example
Find the derivative matrix for w = f(x,y,z) = ( x2 + xy4eyz , ln(xz+y) ) .
_
_
|
|
| 2x + y4eyz
|
4xy3eyz + xy4zeyz
xy5eyz
|
|
|
|
Df(x,y,z) = |
|
|
|
1 / (xz+y)
x / (xz+y)
| z / (xz+y)
|
|
|
|
|
|_
_|
Find the linear approximation at the point x0 = (1 , e , 0).
x–1
w = f (x0) + Df (x0) (x – x0) = f (1 , e , 0) + Df (1 , e , 0) y – e
z–0
=
2 + e4
1+
e4
e5
x–1
y–e
+
1
0
1/e
1/e
=
z–0
(2 + e4)x – 2 – e4 + 4e3y – 4e4 + e5z
1 + e4
1
4e3
+
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,
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(xz+y) is
w2 = (1/e)y + (1/e)z .
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.