Review of Partial Derivatives - Bruce E. Shapiro

Review of Partial Derivatives
Definition of Partial Derivatives. Let z = f( x, y)be a function. Then the partial
derivative of f with respect to x is defined as
∂f
∂ ≡ ≡ f x+ h y − f x y
fx( x, y) lim ( , ) ( , )
x
h→0
h
and the partial derivative of f with respect to y is defined as
∂f
∂ ≡ ≡ f x y+ k − f x y
fy( x, y) lim ( , ) ( , )
y
k→0
k
Partial derivatives may be computed algebraically; all of the same rules that applied to
regular derivatives also apply to partial derivatives. The only trick to remember
when taking a partial derivative is hold all other variables (besides the one
we are differentiating with respect to) constant.
2
x
Example. Let f( x, y)= . Then
1 + y
2
x
fx ( x , y )= ∂
x + y = 1 ∂
+ y x x 2 2
=
∂ 1 1 ∂ 1 +
xy
2
x
x
f y ( x , y ) ∂
= x x y
y + y
= 2 ∂ 1
y + y
= 2 ∂
y
( + ) −1
1 = − ∂ 1 ∂ 1 ∂
( 1 + y )
Example. Find the partial derivatives of f( x, y)= y 2 e 3x
∂
f x y
x ye y ∂
x( , )= ( )= (
x e )= y ⋅ 3 e = 3
∂
∂
y e
∂
f x y
y ye 2 3x e 3x ∂
y
y y 2 e 3x y ye 3x
( , )= ( )= ( )= ⋅ 2 = 2
∂
∂
2 3x 2 3x 2 3x 2 3x
Example. Find the partial derivative of f( x, y, z)=
fx ( x , y )=
∂ ⎛
∂x
⎜
⎝
fy ( x , y )= ∂ ⎛
∂y
⎜
⎝
fz ( x , y )= ∂ ⎛
∂z
⎜
⎝
2 3 3
x y
z
x y
z
2 3
x y
z
3 3
⎞ y
z x x y xy
⎟ x
⎠
= ∂ 2
2
( )= ⋅ 2 =
∂ z z
2 3 2
2 3
x y
z
2
2 2
⎞ xz y y x x y
⎟ y
⎠
= ∂ 3
2 3
( )= ⋅ 3 =
∂ z z
⎞
x y x y
z z z z x y z x y
⎟
⎠
= 2 3 ∂ ⎛ 1⎞
⎝ ⎠ = 2 3 ∂ −1 2 3 −2
( )= ( − )=−
2
∂
∂
z
2
2
2 3
Page 1

Second Order Partial Derivatives
We define higher order partial derivatives in much the same way as we did in single-variable
calculus.
f f
fxx
= ∂ 2
= ∂ ⎛ ∂ ⎞
2 ⎜ ⎟
∂x
∂x
⎝ ∂x⎠
2
f f
fyy
= ∂ = ∂ ⎛ ∂ ⎞
2 ⎜ ⎟
∂y
∂y
⎝ ∂y⎠
With partial derivatives, we can also combine the variables, so there are more derivatives at
each order. For example, we can differentiate f x with respect to y and we can differentiate
f y with respect to x:
f
xy
2
f f
= ∂ ⎛ ∂ ⎞
⎜ ⎟ = ∂ ∂x
⎝ ∂y⎠
∂∂ xy
2
f f
fyx
= ∂ ⎛ ∂ ⎞
⎜ ⎟ = ∂ ∂y
⎝ ∂x⎠
∂∂ yx
We can, of course, combine higher order derivatives in any order we like. For example:
f f
fxxyx = ∂ 4
∂ ∂ ∂
∂x ∂x ∂y
∂ x
= ∂
2
∂ xyx ∂ ∂
There are not as many partials as you might think, however, because of the following
theorem:
f f
The order of the partials can be reversed: fxy
= ∂ 2
fyx
∂∂ xy
= ∂ 2
∂∂ yx
=
Example. Find all the second-order partial derivatives of f( x, y)= xy 2 + 3 x 2 e y and show
that fxy
= fyx
by taking partials in both orders.
2 y
y
y
y
fx
= y + xe ⇒ fxx
= e fyx
= ∂ 2
6 6 , ( y + 6xe )= 2y+
6xe
∂y 2 y
2 y
y
y
fy
= xy+ x e ⇒ fyy
= x+ x e fxy
= ∂ 2
2 3 2 3 , ( 2xy + 3x e )= 2y + 6xe = f
∂x
Example. Repeat the above example for f( x, y)=
xe y
f = e ⇒ f = 0,
f = e
x
y
xx
f = xe ⇒ f = xe , f = e = f
y
y
yy
y
yx
xy
y
y
yx
yx
Page 2

Chain Rule
If zt () = f( xt (), yt ()) then the chain rule is
dz f dx f dy
= ∂ + ∂ dt ∂x
dt ∂y
dt
In general, if z is a function of any number of variables xt ( ), yt ( ), zt ( ), wt ( ), ... , each of
which can be expressed as a function of only t (and not of any other parameter),
d
dt f x y z w f dx f dy f dz f dw
( , , , ,... )= ∂ + ∂ + ∂ + ∂ +
∂x
dt ∂y
dt ∂z
dt ∂w
dt
...
Example. Suppose f( x, y) = xsin
y, where x = t
2 and y = 2t
+ 1. Then by the chain rule
we have
′ = ∂ f dx
+ ∂ f dy
f () t
∂x
dt ∂y
dt
= ∂ ∂ ( ) ( )+ ∂ x x sin y d dt t 2
∂ y ( x sin y) d dt
( 2t
+ 1 )
= ( sin y)( 2t)+ ( xcos
y)( 2)
= 2tsin( 2t + 1)+ 2t 2
cos( 2t
+ 1)
Page 3

Vector Products
There are two types of products between vectors, one of which produces a vector and the
other produces a scalar
• The dot product v r ⋅w
r ⎯→⎯
scalar
• The cross product v r × w r ⎯→⎯
vector
The Dot Product is defined geometrically
r r
v⋅ w = v wcosθ
where q is the angle between the two vectors as shown in the
figure. Algebraically, if
r r r r r r r r
v = iv1+ jv2 + kv3 and w = iw1+ jw2 + kw3
r r r r
Then v⋅ w = w⋅ v = v1w1+ v2w2 + v3w3
Example. Suppose that r r r r
u = 3i + 4 j + 5 k and r r r r
v = 7i + 8j + 9k
Then u r ⋅ v
r = ( 3)( 7) + ( 4)( 8) + ( 5)( 9)
= 21 + 32 + 45 = 98
Properties of the dot product
r r r r
1. v⋅ w = w⋅v
r r r r r r
2. v⋅ ( aw) = ( av) ⋅ w = a( v⋅w)
3. ( v r + u r ) ⋅ w r = v r ⋅ w r + u r ⋅w
r
r
4. v and ware r perpendicular only if v r ⋅ w
r = 0.
The Cross Product
The cross product is a product between vectors that results in a vector. It is defined as a
vector with the following properties:
• Its length is equal to v r × w r = v r w
r sinθ
• direction is perpendicular to the plane that contains v and w
• Its orientation (up vs. down) is according to the right hand rule
Right-Hand Rule: Place u r and v r so that their tails coincide and curl the fingers of your right
hand from through the angle from u r to v r . Your thumb is pointing in the direction of u
r × v
r
The cross product gives the area of the parallelogram formed by the two vectors:
θ
θ
w|sinθ
v|
Page 4

We can also calculate the cross product algebraically from the components of the individual
vectors if we use determinants.
r r r
i j k
r r
r r r
v w v v v i v 2 v 3
j v 1 v 3
k v 1 v 3
× = 1 2 3 = − +
w2 w3
w1 w3
w1 w3
w1 w2 w3
r r r
= i( v w −v w )− j( v w −v w )+ k( v w −v w )
2 3 3 2 1 3 3 1 1 2 2 1
Determinant of a Matrix
⎛
det a b ⎞ a b
⎜ ⎟ ≡ ≡ ad −bc
⎝ c d⎠
c d
⎛ a b c⎞
a b c
det⎜d e f⎟
d e f a e f b d f c d e
⎜ ⎟ = = h i
− g i
+ g h
⎝ g h i⎠
g h i
= a( ei − fh) −b( di − fg) + c( dh −eg)
Properties of the Cross Product
(1) w r × v r = − v r × w
r
r r r r r r
(2) ( av)× w = a( v × w)= v ×( aw)
(3) u r × ( v r + w r )= u r × v r + u r × w
r
(4) u
r × v
r = 0 if and only if the vectors are parallel (assuming that r r r
uv , ≠ 0)
(5) i r × r j = k r , r j × k r = i r , k r × i r =
r
j (cyclic cross products)
(6) r j × i r = − k r , k r × r j = − i r , i r × k = −
r
j (acyclic cross products)
(7) v
r × v
r = 0
Vector Operators: Gradient, Divergence and Curl
The gradient of a function of three variables is the vector
r f f
grad f ≡∇f( x, y, z)
≡i
∂ j k
f ∂ x
+ r ∂ ∂ y
+ r ∂ ∂z
We can think of the gradient symbol ∇ as a operator that behaves exactly like a vector with
the exception of the fact that it operates on the single item immediately to its right.
In this sense, we can think of ∇ as vector version of ∂ ∂x .
Specifically, ∇ represents the “vector”
r r ∂ r r
∇=
∂ + ∂
∂ + ∂
i j k
x y ∂ z
Page 5

Example. f xe y 2
+
=
z
Then
r r r r r
∇ =∇ ⎛ 2 ⎞
⎝ ⎠ = ⎛ ∂
∂ + ∂
∂ + ∂ ⎞
⎜
⎟ ⎛ 2
f xe i j k xe
⎝ x y ∂ z ⎠⎝
y + z y + z
r ∂ ⎛ ⎞ r r
=
∂ ⎝ ⎠ + ∂ ⎛ ⎞
∂ ⎝ ⎠ + ∂ ⎛
i
x xe 2 j y xe 2 k ∂z ⎝
xe 2
r r r
y
2
+ z y
2
+ z y
2
+ z
= i e + j 2xye + kxe
y + z y + z y + z
Since we are treating ∇ “like a vector,” we can subject to all of the usual vector operations,
such as dot product and cross product. We will define the following operations using ∇
Operation Name of Operator Input Output
∇ gradient scalar vector
∇⋅ divergence (dot product) vector scalar
read as “del dot ...”
∇× curl (cross product) vector vector
read as “del cross ...”
∇⋅∇ or ∇ 2 Laplacian scalar scalar
read as “del squared ...”
Both the divergence and curl operate on vector functions. A vector function is a function
r r r r
Fxyz ( , , ) = iF( xyz , , ) + jF( xyz , , ) + kF( xyz , , )
1 2 3
where the functions F1, F2, F3
are all real valued functions.
Example of a Vector Function. r r 2
Fxyz ( , , )= xi+ yx+
zj r ek
x r
⎞
⎠
( ) +
The Divergence Operator is defined as “del dot a vector function”,
div r r r ⎛ r ∂
F =∇⋅ F = i
r j k r iF r r jF kF
r
∂ x
+ ∂
∂ y
+ ∂ ⎞
⎜
⎟ ⋅ z
( 1+ 2 + 3)
⎝
∂ ⎠
F F F
= ∂ 1
∂ x
+ ∂ 2
∂ y
+ ∂ 3
∂z
Example. Find the divergence of r r 2
Fxyz ( , , )= xi+ yx+
zj r ek
x r
⎞
⎠
( ) +
( )
r r ⎛ r ∂ r r r r r
∇⋅ =
∂ + ∂
∂ + ∂ ⎞
2
x
F ⎜i
j k ⎟ ⋅ xi + ( y x + z) j + e k
⎝ x y ∂ z ⎠
= ∂ x
∂ + ∂ 2
( + )+ ∂ = + + = +
x ∂y y x z x
e 1 2xy 0 1 2xy
∂z Page 6

The Curl Operator is defined as “del cross f”
curl r r r ⎛ r ∂
F =∇× F = i
r j k r iF r r jF kF
r
∂ x
+ ∂
∂ y
+ ∂ ⎞
⎜
⎟ × z
( 1+ 2 + 3)
⎝
∂ ⎠
r r r
i j k
r F F r F F r F
= ∂ ∂ ∂
i
j
k
∂x ∂y ∂ z
= ⎛ ∂
∂y
− ∂ ⎞ ⎛ ∂
⎜ ⎟ +
⎝ ∂z
⎠ ∂z
− ∂ ⎞ ⎛ ∂
⎜ ⎟ +
⎝ ∂x
⎠
⎜
⎝ ∂x
F F F
1 2 3
Example. Suppose F r e y r y
= 2 i + 2 xye 2r
j + k
r
r r r
i j k
r r
∇× = ∂ ∂ ∂
F
∂x ∂y ∂z
y
2
y
2
e 2xye
1
− ∂ F ⎞
⎟
∂y
⎠
3 2 1 3 2 1
r⎛
∂
r r
= − ∂ ⎞ ∂
⎜
⎟ + − ∂ ⎛
⎛
⎞ ∂
⎜
⎟ +
− ∂ y
() 1
y
2
i
xye j
⎝ ∂y
∂z
⎠ ⎝ ∂z e y
2 () 1
y
2 e
[ 2 ]
k [ 2xye
]
∂x
⎠ ⎜
⎝
∂x
∂y
r r r
= i( 0−0)+ j( 0−0)+ ⎛ y y
k 2ye 2 −2ye
2 ⎞
⎝
⎠ = 0
The Laplacian Operator is the dot product of ∇with itself, or the divergence of the
gradient. It is sometimes read as “del squared”.
∇
2 r r
f( x, y, z) =∇⋅ ∇f( x, y, z)
( )
⎛ r ∂
=
∂ + r ∂
∂ +
r ∂ ⎞ ⎡⎛
r ∂
⎜
⎟ ⋅
⎝
∂ ⎠ ∂ + r ∂
∂ +
r ∂ ⎞ ⎤
i j k ⎢⎜i
j k ⎟ f ⎥
x y z ⎣⎝
x y ∂ z ⎠ ⎦
⎛ r ∂
=
∂ + r ∂
∂ +
r ∂ ⎞ ⎛ r∂f
⎜i
j k ⎟ ⋅ i
⎝
∂ ⎠ ∂ + r
j
∂ f
x y z x ∂ +
r ∂ ⎞
⎜
k
f ⎟
⎝ y ∂z⎠
2
= ∂ f
+ ∂ f
+ ∂ f
2 2 2
∂x
∂y
∂z
The Laplacian operates on a scalar and its output is a scalar.
2
2
2
⎞
⎟
⎠
Page 7

Polar Coordinates
In polar coordinates, instead of using the distances x and y from the origin to locate a point,
we use a single distance r and an orientation angle with respect to the x-axis.
x=rcosθ
r
y=rsinθ
θ
The two coordinate systems are related to each other as follows:
x = rcosθ
r = x
2
+ y
2
y = rsinθ
y
θ = arctan
x
We can also define unit vectors and r θ at any given point in space; they can be related to
the cartesian unit vectors i r and r j from the following geometry.
r
j
r
θ
θ
θ
r
r
i
P
θ
Since all the vectors shown are unit vectors, we observe that:
x component of ˆr is cosθ ;
y component of ˆr is sinθ ;
x component of ˆθ is −sinθ
y component of ˆθ is cosθ
Page 8

Therefore
rˆ = cosθi ˆ + sinθj
ˆ
θˆ =− sinθi
ˆ + cosθj
ˆ
Exercise: Find expressions for î and ĵ in terms of ˆr and ˆθ .
Solution. Multiply the first expression by sinθ and the second expression by cosθ
ˆsin sin cos ˆ 2
r θ = θ θi + sin θjˆ
θˆ cos θ cos θ sin θiˆ 2
=− + cos θjˆ
Adding the two expressions,
ˆsin ˆ cos sin
2 ˆ 2
r θ + θ θ = θj + cos θjˆ = jˆ
which gives an expression for ĵ in terms of ˆr and ˆθ .
To get a similar expressionf or î , multiply the expression for ˆr by cosθ and the
expression for ˆθ by sinθ ,
2
rˆ cosθ = cos θiˆ + cosθsinθj
ˆ
ˆsin sin
2
θ θ =− θiˆ + cosθsinθj
ˆ
Subtract the bottom expression from the top,
ˆ cos ˆsin cos
2 ˆ 2
r θ − θ θ = θi + sin θiˆ = iˆ
Summarizing the conversion expressions, we have
r r
i = rˆ cosθ −θˆsin
θ r = cosθi + sinθj
and r r r
jˆ = rˆsinθ + θˆ cos θ θ =− sinθi
+ cosθj
Example. Find an expression for the vector field F r = 3xiˆ+
4x 2 yjˆ in polar coordinates.
Solution.
r
F = 3xiˆ
2
+ 4x yjˆ
= 3( rcos θ)(ˆ rcosθ − θˆsin θ ) + 4 ( r cos θ ) ( r sin θ )(ˆ r cos θ + θˆsin θ )
2 3 3 3 2 2
= 3rcos θrˆ − 3rcosθsinθθ ˆ + 4r cos θsinθrˆ + 4r
cos θsin θθˆ
2 3 3
3 2 2
= ( 3rcos θ + 4r cos θsin θ)ˆ r + ( 4r
cos θsin θ − 3rcosθsin θ) θ ˆ
2 2 2
= rcos θ( 3+ 4r cosθsin θ)ˆ r + rcosθsin θ( 4r
cosθsin θ −3) θ ˆ
Exercise: Recall that the gradient in two dimensions in cartesian coordinates is
∇ = ∂ f
f x y i
∂ + ∂ f
( , ) ˆ jˆ . Find an expression for the gradient in polar coordinates, i .e., find
x ∂y
the functions g and h so that ∇ f(, r θ ) = rg ˆ (, r θ ) + θ ˆ h (, r θ ).
Hint: you must use the chain
rule and the expressions for transformation of coordinates.
2
Page 9