Math 314 Lecture #13 §14.3: Partial Derivatives

Math 314 Lecture #13
§14.3: Partial Derivatives
Outcome A: Find the first partial derivatives of a function of several variables.
By choosing one of the independent variables x, y in a function of two variables f (x, y) to
be a constant, we get a function of just one variable to which we can apply the familiar
notion of slope and rules of differentiation.
The partial derivative of a function f (x, y) with respect to x at the point (a, b) is
f (a + h, b) − f (a, b)
h→0
h
fx (a, b) = lim
when the limit exist. Here is a picture of the intersection of z = f (x, y) with y = b.
The partial derivative of a function f (x, y) with respect to y at the point (a, b) is
f (a, b + h) − f (a, b)
h→0
h
fy (a, b) = lim
when the limit exist. Here is a picture of the intersection of z = f (x, y) with x = a.
We also have the first-order partial derivative functions:
f (x + h, y) − f (x, y)
,
h→0
h
f (x, y + h) − f (x, y)
fy (x, y) = lim
.
h→0
h
fx (x, y) = lim
There are various notations for these first-order partials of z = f (x, y):
∂f
∂z
=
= zx = D1 f
∂x
∂x
∂f
∂z
fy =
=
= zy = D2 f.
∂y
∂y
fx =
The symbol ∂ in these notations is read “del,” and we read Di as the partial derivative
with respect to the ith variable.
For functions of three or more variables, we hold constant all but one variable and
differentiate with respect to that variable to get the first-order partial derivative with
respect to that variable at a point.
ey
, we have
Examples. (a) For z = f (x, y) =
x + y2
y ∂
e
−ey
fx =
,
=
∂x x + y 2
(x + y 2 )2
y ∂
e
ey (x + y 2 ) − ey (2y)
fy =
.
=
∂y x + y 2
(x + y 2 )2
(b) For w = f (x, y, z) = x sin(y − z), we have
fx = sin(y − z), fy = x cos(y − z), fz = −x cos(y − z).
Outcome B: Find the second and higher order partial derivatives of a function of several
variables.
The second-order partials of z = f (x, y) are the first-order partials of the f :
∂2f
∂2z
∂ ∂f
=
fxx = (fx )x =
=
= D1 D1 f = zxx ,
∂x ∂x
∂x2
∂x2
∂2f
∂2z
∂ ∂f
fxy =
=
=
= D2 D1 f = zxy ,
∂y ∂x
∂y∂x
∂y∂x
∂ ∂f
∂2f
∂2z
fyx =
=
=
= D1 D2 f = zyx ,
∂x ∂y
∂x∂y
∂x∂y
∂ ∂f
∂2f
∂2z
fyy = (fx )x =
= 2 = 2 = D2 D2 f = zyy .
∂y ∂y
∂y
∂y
For a function w = f (x, y, z), there are 9 second-order partial derivaties: fxx , fxy , fxz ,
fyx , fyy , fyz , fzx , fzy , and fzz .
Third-order partial derivatives of z = f (x, y) are computed by partial differentiation of
the second-order partial derivatives: fxxx , fxxy , etc.
p
Examples. (a) For z = f (x, y) = x2 + y 2 , we have
x
zx = p
x2 + y 2
, zy = p
y
x2 + y 2
,
and so
zxx =
zxy =
zyx =
zyy =
p
p
x2 + y 2 − x(1/2)(x2 + y 2 )−1/2 (2x)
x2 + y 2 − x2 (x2 + y 2 )−1/2
=
,
x2 + y 2
x2 + y 2
−x(1/2)(x2 + y 2 )−1/2 (2y)
xy
=− 2
,
2
2
x +y
(x + y 2 )3/2
xy
−y(1/2)(x2 + y 2 )−1/2 )2x)
=
−
,
x2 + y 2
(x2 + y 2 )3/2
p
p
x2 + y 2 − y(1/2)(x2 + y 2 )−1/2 (2y)
x2 + y 2 − y 2 (x2 + y 2 )−1/2
=
.
x2 + y 2
x2 + y 2
It is just a coincidence that zxy = zyx ? That the so-called mixed second order partial
derivatives are the same? Maybe this is lucky because of the form of f ?
(b) For z = xy 2 − x3 , we have
zx = y 2 − 3x2 , zy = 2xy,
and so
zxx = 6x, zxy = 2y, zyx = 2y, zyy = 2x.
Again, we see that the mixed second-order partials agree!
Outcome C: Verify the conclusion of Clairaut’s Theorem.
Theorem. Suppose z = f (x, y) is defined on a disk D that contains the point (a, b). If
fxy and fyx are both continuous on D, then fxy = fyx .
A similar result holds for functions f (x, y, z) of three or more variables, i.e., fxy = fyx ,
fxz = fzx , and fyz = fzy , under the appropriate continuity conditions.
Clairaut’s Theorem also extends to third and higher order partial derivatives under the
appropriate continuity conditions: fxxy = fxyx = fyxx , etc.
Example. For z = f (x, y) = x sin(x + 2y) we have
zx = sin(x + 2y) + x cos(x + 2y), zy = 2x cos(x + 2y),
and so
zxy = 2 cos(x + 2y) − 2x sin(x + 2y) = zyx = 2 cos(x + 2y) − 2x sin(x + 2y).

Download Report