PARTIAL DERIVATIVES
The rules that apply to functions of 1 variable
also to functions of 2 or more variables.
However, a function of 2 variables can change
in response to a change in either of its
variables. If one of its variables is fixed and
the other changed infinitesimally, the ratio of
the resultant infinitesimal change in the
function to the change in that variable is the
PARTIAL DERIVATIVE of the function with
respect to that variable.
z = f ( x , y)
f x( x0 , y0) =
f y( x0 , y0) =
lim
x → x0
lim
y → y0
f ( x , y0) − f ( x0 , y0)
x − x0
f ( x 0 , y ) − f ( x 0 , y 0)
y − y0
Page 1 of 14
Equivalent Notations:
f x( x0 , y0) =
∂
∂x
f ( x0 , y0) f y( x0 , y0) =
∂
∂y
f ( x 0 , y 0)
Higher order partial derivatives can also be
computed for multivariable functions just like
single variable functions.
Example # 1: Find the first and ALL of the
second partial derivatives of the given
function.
2
f ( x , y) = x ⋅ y
2
Here are the First Partial Derivatives.
∂
∂x
f = 2⋅ x⋅ y
2
∂
∂y
2
f = 2⋅ x ⋅ y
Here are the Second Partial Derivatives.
Page 2 of 14
⎞
∂2
2
=
=
2
⋅
y
f
f
⎜ ⎟
2
∂x ⎝ ∂x ⎠
∂x
∂ ⎛∂
⎞ ∂ ∂
f = 4⋅ x⋅ y
⎜ f⎟ =
∂y ⎝ ∂x ⎠ ∂y ∂x
∂ ⎛∂
⎞ ∂ ∂
f = 4⋅ x⋅ y
⎜ f⎟ =
∂x ⎝ ∂y ⎠ ∂x ∂y
∂ ⎛∂
⎞
∂2
2
f
=
=
2
⋅
x
f
⎜ ⎟
2
∂y ⎝ ∂y ⎠
∂y
∂ ⎛∂
Note that in this case that the Mixed Second
Partial Derivatives are equal. They are always
equal if they are both continuous functions.
∂ ∂
∂x ∂y
f =
∂ ∂
∂y ∂x
f
Page 3 of 14
Definition: Let " z = f ( x , y) " be differentiable
at the point: ( x0 , y0). Denote a new function,
" dz = f x( x0 , y0) ⋅ dx + f y( x0 , y0) ⋅ dy " with
dependent variable " dz " and independent
variables:
" dx " and " dy ". This "new" function is
refered to as the Total Differential.
Example # 2: Compute the total differential of
the given function at the point: ( 1 , 2 , 4).
2
z = f ( x , y) = x ⋅ y
2
⎛∂ ⎞
⎛∂ ⎞
dz = ⎜ f ⎟ ⋅ dx + ⎜ f ⎟ ⋅ dy
⎝ ∂x ⎠
⎝ ∂y ⎠
∂
∂x
f = 2⋅ x⋅ y
(
2
)
2
d
2
f = 2⋅ x ⋅ y
dy
(
2
)
dz = 2 ⋅ x ⋅ y ⋅ dx + 2 ⋅ x ⋅ y ⋅ dy
2
2
dz = ⎡⎣ 2 ⋅ ( 1) ⋅ ( 2) ⎤⎦ ⋅ dx + ⎡⎣2 ⋅ ( 1) ⋅ ( 2)⎤⎦ ⋅ dy
dz = ( 8) ⋅ dx + ( 4) ⋅ dy
Page 4 of 14
Example # 3: Compute the total differential of
the given function at the arbitrary point:
( x , y , f ( x , y) ).
z = f ( x , y) = tan
− 1(
x ⋅ y)
⎛∂ ⎞
⎛∂ ⎞
dz = ⎜ f ⎟ ⋅ dx + ⎜ f ⎟ ⋅ dy
⎝ ∂x ⎠
⎝ ∂y ⎠
∂
∂x
∂
∂y
f = ⎛⎜
⎞⋅ y
⎟
+
⋅
1
x
y
⎝
⎠
1
f = ⎛⎜
1
⎞⋅⎛ x
⎟⎜
+
⋅
1
x
y
⎝
⎠ ⎝ 2⋅
⎛
y
⎞
⎟
y⎠
⎞
x
⎡
⎤ ⋅ dy
⋅
dx
+
⎟
⎢
⎥
(
)
1
+
x
⋅
y
⋅
⋅
1
+
x
⋅
y
2
y
⎝
⎠
⎣
⎦
dz = ⎜
The problems that follow illustrate
application of the total differential.
Page 5 of 14
Example # 4: A point moves along the
intersection of the given elliptic paraboloid
and plane. At what rate is "z " changing with
"x" when the point is at ( 1 , 1 , 4) ? Also, find
the equation of the tangent line at that point.
2
z ( x , y) = x + 3 ⋅ y
2
y=1
Paraboloid & Plane
z
y
x
2
z ( x , y) = x + 3 ⋅ y
2
y=1
The two surfaces intersect here.
2
z ( x , 1) = x + 3
Page 6 of 14
dz = 2 ⋅ x ⋅ dx
dz = 2 ⋅ ( 1) ⋅ dx
dy = 0
Tangent Line equations are thus these.
x ( t ) = 1 + ( 1) ⋅ t y ( t ) = 1 + ( 0) ⋅ t
z ( t ) = 4 + ( 2) ⋅ t
⎡ 1 + ( 1) ⋅ t ⎤
⎯⎯→ ⎢
⎥
LT ( t) = ⎢ 1 + ( 0) ⋅ t ⎥
⎢ 4 + ( 2) ⋅ t ⎥
⎣
⎦
Those are the parametric form for the
equation of the tangent line at the given
point on the curve defined by the
intersection of the elliptic paraboloid and
the plane.
Page 7 of 14
Tangent@ (1,1,4)
z
y
x
Example # 5: Find the equation of the tangent
line at the point : ( 1, 1, 4 ) on the curve
defined by the intersection of the given
elliptic paraboloid and plane.
2
z1 ( x , y) = x + 3 ⋅ y
2
z2 ( x , y) = 2 ⋅ x + 2 ⋅ y
Page 8 of 14
Parabaloid & Tilted Plane
z
x
y
dz1 = 2 ⋅ x ⋅ dx + 6 ⋅ y ⋅ dy
2
dz2 = 2 ⋅ dx + 2 ⋅ dy
2
x + 3⋅ y = 2⋅ x + 2⋅ y
⎛d ⎞
d
1 + 1⋅ y = 1⋅ x + 3⋅ y⋅ ⎜ y⎟
dx
⎝ dx ⎠
Page 9 of 14
⎛d ⎞
⎜ y ⎟ ⋅ ( 3 ⋅ y − 1) = 1 − x
⎝ dx ⎠
1−x
d
y=
3⋅ y − 1
dx
1 − ( 1)
d
y=
=0
3 ⋅ ( 1) − 1
dx
dy = ( 0)dx
dz1 = 2 ⋅ x ⋅ dx + 6 ⋅ y ⋅ dy
dz1 = 2 ⋅ x ⋅ dx + 6 ⋅ y ⋅ ( 0) ⋅ dx
dz1 = 2 ⋅ ( 1) ⋅ dx
dz2 = 2 ⋅ dx + 6 ⋅ dy
dz2 = 2 ⋅ dx + 6 ⋅ ( 0) ⋅ dx
dz2 = 2 ⋅ dx
dz1 = dz2 = dz = ( 2) ⋅ dx
⎡ 1 + ( 1) ⋅ t ⎤
⎯⎯→ ⎢
⎥
LT ( t) = ⎢ 1 + ( 0) ⋅ t ⎥
⎢ 4 + ( 2) ⋅ t ⎥
⎣
⎦
Page 10 of 14
Tangent@ (1,1,4)
z
x
y
Example # 6: Find the equation of the tangent
line at the point:" ( 1 , 0 , 1)" on the curve
defined by the equation of the intersection of
the given surfaces.
Page 11 of 14
2
z1 ( x , y) = 2 − x − y
2
2
z2 ( x , y) = x + y
2 Curved Surfaces
z
y
x
2
z1 ( x , y) = 2 − x − y
2
2
2
z2 ( x , y) = x + y
2
2
2−x −y = x +y
−2 ⋅ x − 2 ⋅ y ⋅
d
d
y = 2⋅ x + y
dx
dx
Page 12 of 14
( 1 + 2 ⋅ y) ⋅
d
y = −4 ⋅ x
dx
−4 ⋅ x
d
y=
1 + 2⋅ y
dx
Substitute the values of the "x" and "y"
coordinates of the given point: ( 1 , 0 , 1).
−4 ⋅ ( 1)
d
y=
= −4
1 + 2 ⋅ ( 0)
dx
dy = −4 ⋅ dx
dz1 = −2 ⋅ x ⋅ dx − 2 ⋅ y ⋅ dy
dz2 = 2 ⋅ x + 1 ⋅ dy
dx==−12⋅⋅ dx
dz1
x ⋅ dx − 2 ⋅ y ⋅ ( −4 ⋅ dx) dz2 = 2 ⋅ x + 1 ⋅ ( −4 ⋅ dx)
dz1 = ( −2 ⋅ x + 8 ⋅ y) ⋅ dx
dz2 = ( 2 ⋅ x − 4) ⋅ dx
Page 13 of 14
dz1 = [ −2 ⋅ ( 1) + 8 ⋅ ( 0) ] ⋅ dx
dz2 = [ 2 ⋅ ( 1) − 4] ⋅ dx
dz1 = dz2 = dz = ( −2) ⋅ dx
⎡ 1 + ( 1) ⋅ t ⎤
⎯⎯→ ⎢
⎥
LT ( t) = ⎢ 0 + ( −4) ⋅ t ⎥
⎢ 1 + ( −2) ⋅ t ⎥
⎣
⎦
Tangent@(1,0,1)
z
y
x
Page 14 of 14