Calculus 3
Lia Vas
Partial Derivatives
Let z = f (x, y) be a function of two variables. The partial derivative of z with respect to x
is obtained by regarding y as a constant and differentiating z with respect to x. Notation:
∂z
.
∂x
The derivative zx at a point (x0 , y0 , z0 ) on the
surface z = f (x, y), represent the rate of change
of function z = f (x, y0 ) in the direction of xaxis. This value is the slope of the line tangent to
the curve obtained by intersection of the surface
z = f (x, y) and the vertical plane y = y0 .
zx
or
The partial derivative of z with respect to y is obtained by regarding x as a constant and
differentiating z with respect to y. Notation:
∂z
.
∂y
This derivative at a point (x0 , y0 , z0 ) on the surface z = f (x, y), representthe rate of change of
function z = f (x0 , y) in the direction of x-axis.
This value is the slope of the line tangent to
the curve obtained by intersection of the surface
z = f (x, y) and the vertical plane x = x0 .
zy
or
The second partial derivatives can be found by differentiating the first partial derivatives.
Differentiating zx with respect to x produces
(zx )x = zxx
∂
=
∂x
∂z
∂x
!
=
∂ 2z
∂x2
and with respect to y produces
∂z
∂x
!
∂z
∂y
!
(zy )x = zyx
∂
=
∂x
(zy )y = zyy
∂
=
∂y
∂z
∂y
(zx )y = zxy
∂
=
∂y
=
∂ 2z
.
∂y∂x
=
∂ 2z
∂x∂y
Differentiating zy with respect to x produces
and with respect to y produces
!
∂ 2z
= 2.
∂y
If zxy and zyx are both continuous, then
zxy = zyx .
In this case, differentiating zx with respect to y and zy with respect to x produces the same result.
Practice problems. Find the first and the second partial derivatives.
1. z = 3x2 + 2xy − 5y 2
2. z = ex sin y
2 −xy
3. z = ex
5. z = ax2 ex
4. z = x ln(xy 2 )
2 −xy
where a is a constant (“PChem problem”).
6. Assuming that a is also a variable in the previous problem, find
∂z
.
∂a
Solutions.
1. zx = 6x + 2y, zy = 2x − 10y, zxx = 6, zxy = zyx = 2, zyy = −10.
2. zx = ex sin y, zy = ex cos y, zxx = ex sin y, zxy = zyx = ex cos y, zyy = −ex sin y.
2
2
3. zx = ex −xy (2x −y), zy = ex −xy (−x), zxx = ex
2
2
y)(−x) − ex −xy , zyy = ex −xy x2 .
2
2 −xy
y
2xy
2
4. zx = ln(xy 2 ) + x xy
2 = ln(xy ) + 1, zy = x xy 2 =
2
(2x−y)2 +ex
2x
,
y
zxx =
y2
xy 2
2
5. zx = 2axex −xy + ax2 ex −xy (2x − y) = a(2x + 2x3 − x2 y)ex
2
2
zy = ax2 ex −xy (−x) = −ax3 ex −xy .
2
Then zxx = a(2 + 6x2 − 2xy)ex −xy + a(2x + 2x3 − x2 y)ex
2
2
zyy = −ax3 ex −xy (−x) = ax4 ex −xy .
2 −xy
(2), zxy = zyx = ex
2 −xy
= x1 , zxy = zyx = y2 , zyy =
2 −xy
2 −xy
(2x−
−2x
.
y2
,
(2x − y) and
2
To find zxy , either differentiate zx with respect to y and get zxy = a(−x2 )ex −xy + a(2x + 2x3 −
2
2
2
x2 y)ex −xy (−x) = a(−x2 − 2x2 − 2x4 + x3 y)ex −xy = a(−3x2 − 2x4 + x3 y)ex −xy
or differentiate zy with respect to x and get zxy = −3ax2 ex
2
2x4 + x3 y)ex −xy
6. z = ax2 ex
2 −xy
⇒
∂z
∂a
= x2 e x
2 −xy
2 −xy
− ax3 ex
2 −xy
(2x − y) = a(−3x2 −