14.3: Partial Derivatives,
14.4: The Chain Rule, and
14.5: Directional Derivatives and Gradient Vectors
TA: Sam Fleischer
November 10
Section 14.3: Partial Derivatives
Definition: Partial Derivative with respect to x, y
The partial derivative of f (x, y) with respect to x at the point (x0 , y0 ) is
∂f d
f (x0 + h, y0 ) − f (x0 , y0 )
fx =
=
= lim
f
(x,
y
)
0
h→0
∂x (x0 ,y0 )
dx
h
(x0 ,y0 )
The partial derivative of f (x, y) with respect to y at the point (x0 , y0 ) is
d
∂f f (x0 , y0 + h) − f (x0 , y0 )
=
fy =
f (x0 , x)
= lim
h→0
∂y (x0 ,y0 )
dy
h
(x0 ,y0 )
Example: Taking Partial Derivatives
f (x, y) = y sin (xy)
To find fx (the partial derivative of f with respect to x), treat y as a constant.
fx = y cos (xy)y = y 2 cos (xy)
To find fy (the partial derivative of f with respect to y), treat x as a constant.
fy = (1) sin (xy) + y cos (xy)x = sin (xy) + xy cos (xy)
Example: Implicit Differentiation
Find
∂z
if the equation
∂x
yz − ln z = x + y
defines z as a function of two independent variables x and y and the partial derivative exists.
d
d
(yz − ln z) =
(x + y)
dx
dx
1
∂z
1 ∂z
y
−
=1+0
z ∂x
∂x 1 ∂z
y−
=1
z ∂x
∂z
x
=
∂x
yz − 1
Definition: Second-Order Partial Derivatives
When we differentiate a function f (x, y) twice, we produce its second-order derivatives. Notation:
∂ 2f
∂ 2f
or
f
,
or fyy
xx
∂x2
∂y 2
∂ 2f
∂ 2f
or fyx , and
= fxy
∂x∂y
∂y∂x
The defining equations are
∂ 2f
∂ ∂f
=
,
∂x2
∂x ∂x
∂ 2f
∂ ∂f
=
∂x∂y
∂x ∂y
∂ 2f
Note that fyx =
means you first take the derivative with respect to y, and then take the
∂x∂y
derivate with respect to x.
In general, fxy 6= fyx .
Theorem 2: Mixed Derivative Theorem
If f (x, y) and its partial derivatives fx , fy , fxy , and fyx are defined throughout an open region
containing a point (a, b) and are all continuous at (a, b), then
fxy (a, b) = fyx (a, b)
Example: Partial Derivatives of Higher Order
Find fyxyz if f (x, y, z) = 1 − 2xy 2 z + x2 y
f (x, y, z) = 1 − 2xy 2 z + x2 y
fy = −4xyz + x2
fyx = −4yz + 2x
fyxy = −4z
fyxyz = −4
2
Definition: Differentiability
A function z = f (x, y) is differentiable at (x0 , y) if fx (x0 , y0 ) and fy (x0 , y0 ) exists and ∆z satisfies
an equation of the form
∆z = fx (x0 , y0 )∆x + fy (x0 , y0 )∆y + 1 ∆x + 2 ∆y
in which each of 1 , 2 → 0 as both ∆x, ∆y → 0. We call f differentiable if it is differentiable at
every point in its domain, and say that its graph is a smooth surface.
Theorem 3: The Increment Theorem for Functions of Two Variables
Suppose that the first partial derivatives of f (x, y) are defined throughout an open region R containing the point (x0 , y0 ) and that fx and fy are continuous at (x0 , y0 ). Then the change
∆z = f (x0 + ∆x, y0 + ∆y) − f (x0 , y0 )
in the value of f that results from moving from (x0 , y0 ) to another point (x0 + ∆x, y0 + ∆y) in R
satisfies an equation of the form
∆z = fx (x0 , y0 )∆x + fy (x0 , y0 )∆y + 1 ∆x + 2 ∆y
in which each of 1 , 2 → 0 as both ∆x, ∆y → 0.
Corollary of Theorem 3
If the partial derivatives fx and fy of a function f (xy) are continuous throughout an open region
R, then f is differentiable at every point of R.
Theorem 4 - DIfferentiability Implies Continuity
If a function f (x, y) is differentiable at (x0 , y0 ), then f is continuous at (x0 , y0 ).
Section 14.4: The Chain Rule
Theorem 5: Chain Rule for Functions of One Independent Variable and Two Intermediate Variables
If w = f (x, y) is differentiable and if x = x(t), y = y(t) are differentiable functions of t, then the
composite w = f (x(t), y(t)) is a differentiable function of t and
dw
= fx (x(t), y(t)) · x0 (t) + fy (x(t), y(t)) · y 0 (t)
dt
or
dw
∂f dx ∂f dy
=
+
dt
∂x dt
∂y dt
3
Theorem 6: Chain Rule for Functions of One Independent Variable and Three Intermediate Variables
If w = f (x, y, z) is differentiable and x, y, and z are differentiable functions of t, then w is a
differentiable function of t and
∂w dx ∂w dy ∂w dz
dw
=
+
+
dt
∂x dt
∂y dt
∂z dt
Example
Find
dw
if
dt
w = xy + z,
x = cos t,
y = sin t,
z=t
Use Theorem 6:
dw
∂w dx ∂w dy ∂w dz
=
+
+
dt
∂x dt
∂y dt
∂z dt
= (y)(− sin t) + (x)(cos t) + (1)(1)
= (sin t)(− sin t) + (cos t)(cos t) + 1
= − sin2 t + cos2 t + 1
= cos 2t + 1
What is the derivative at t = 0?
dw = 1 + cos 0 = 2
dt t=0
Theorem 7: Chain Rule for Two Independent Variables and Three Intermediate Variables
Suppose that w = f (x, y, z), x = g(r, s), y = h(r, s), and z = k(r, s). If all four functions are
differentiable, then w has partial derivatives with respect to r and s given by the formulas
∂w
∂w ∂x ∂w ∂y ∂w ∂z
=
+
+
∂r
∂x ∂r
∂y ∂r
∂z ∂r
∂w
∂w ∂x ∂w ∂y ∂w ∂z
=
+
+
∂s
∂x ∂s
∂y ∂s
∂z ∂s
Theorem 8: A Formula for Implicit Differntiation
Suppose that F (x, y) is differentiable and that the equation F (x, y) = 0 defines y as a differentiable
function of x. Then at any point where Fy 6= 0,
dy
Fx
=−
dx
Fy
4
Example
Find
dy
if y 2 − x2 − sin xy = 0.
dx
dy
Fx
=−
dx
Fy
−2x − y cos xy
=−
2y − x cos xy
2x + y cos xy
=
2y − x cos xy
Expansion of Theorem 8 to Three Variables
Suppose F (x, y, z) = 0 and z = f (x, y). Assuming F and f are differentiable functions, then
∂z
Fx
=−
∂x
Fz
and
∂z
Fy
=−
∂y
Fz
Expansion of Chain Rule to Functions of n variables
In general, suppose z = f (x1 , x2 , . . . , xn ) is a differential function of the intermediate variables
x1 , x2 , . . . , xn where n is a positive integer (n ∈ N). Also suppose each xi is a differentiable function
of the independent variables t1 , t2 , . . . , tm , with m ∈ N. In equation form,
x1 = g1 (t1 , t2 , . . . , tm )
x2 = g2 (t1 , t2 , . . . , tm )
..
.
xn = gn (t1 , t2 , . . . , tm )
Then w is a differential function of each of the independent variables t1 , t2 , . . . , tm , and the partial
derivatives of w with respect to each ti are
n
∂w ∂x1
∂w ∂x2
∂w ∂xn X ∂w ∂xi
∂w
=
+
+ ··· +
=
∂t1
∂x1 ∂t1
∂x2 ∂t1
∂xn ∂t1
∂xi ∂t1
i=1
n
∂w
∂w ∂x1
∂w ∂x2
∂w ∂xn X ∂w ∂xi
=
+
+ ··· +
=
∂t2
∂x1 ∂t2
∂x2 ∂t2
∂xn ∂t2
∂xi ∂t2
i=1
..
.
n
∂w
∂w ∂x1
∂w ∂x2
∂w ∂xn X ∂w ∂xi
=
+
+ ··· +
=
∂tm
∂x1 ∂tm ∂x2 ∂tm
∂xn ∂tm
∂xi ∂tm
i=1
More compactly,
n
∂w X ∂w ∂xi
=
∂tj
∂xi ∂tj
i=1
for j = 1, 2, . . . , m
5
Directional Derivatives and Gradient Vectors
Definition: Directional Derivative
The derivative of f at P0 (x0 , y0 ) in the direction of the unit vector u = u1 i = u2 j is the
number
df
f (x0 + su1 , y0 + su2 ) − f (x0 , y0 )
= lim
ds u,P0 s→0
s
provided the limit exists. The directional derivative is also denoted
df
= (Du f )P0
ds u,P0
and is read “The derivative of f at P0 in the direction of u”.
Definition: Gradient Vector
The gradient vector (gradient) of f (x, y) at a point P is the vector
∇f =
∂f
∂f
i+
j
∂x
∂y
Theorem 9: The Directional Derivative is a Dot Product
If f (x, y) is differentiable in an open region containing P0 (x0 , y0 ), then
(Du f )P0 = (∇f )P0 · u
In words, the derivative of f at P0 in the direction of u is the dot product of the gradient ∇f at
P0 and u. In brief,
(Du )f = ∇f · u
Example
Find the derivative of f (x, y) = xey + cos xy at the point (2, 0) in the direction of v = 3i − 4j.
First, find the unit direction vector
u=
v
v
3
4
= = i− j
kvk
5
5
5
Then we need to find the partial derivatives of f at (2, 0) because together, they make up the
gradient, ∇f .
fx = ey − y sin xy
fx (2, 0) = e0 − 0 sin(2 · 0) = 1 − 0 = 1
fy = xey − x sin xy
fy (2, 0) = 2e0 − 2 sin(2 · 0) = 2 − 0 = 2
6
Plug these values into the definition of gradient.
∇f |(2,0) = fx (2, 0)i + fy (2, 0)j
= i + 2j
Then the directional derivative of f at (2, 0) in the direction of u is
(Du f )(2,0) = ∇f |(2,0) · u
3
4
=
i − j · (i + 2j)
5
5
3
4
= −2·
5
5
= −1
Properties of the Directional Derivative
1. The function f increases most rapidly when cos θ = 1, i.e. when θ = 0, i.e. when u is the
direction of ∇f . That is, at each point P in its domain, f increases most rapidly in the
direction of the gradient vector ∇f at P . The derivative in this direction is
Du f = k∇f k cos 0 = k∇f k
2. Similarly, f decreases most rapidly in the direction of −∇f . The derivative in this direction
is
Du f = k∇f k cos π = −k∇f k
3. Any direction u orthogonal to a gradient ∇f =
6 0 is adirection of zero change in f because
π
θ = and
2
π
Du f = k∇f k cos = k∇f k · 0 = 0
2
Example
x2 y 2
Let f (x, y) =
+ , and consider the point (1, 1).
2
2
The function increases most rapidly in the direction of ∇f .
(∇f ) = xi + yj =⇒ (∇f )(1,1) = i + j
The unit vector of (∇f )(1,1) is
i+j
1
1
u= √ = √ i+ √ j
2
2
2
The function decreases most rapidly in the direction −(∇f )(1,1)
1
1
−√ i − √ j
2
2
The directions of zero change at (1, 1) are the directions ofthogonal to ∇f :
1
1
n = −√ i + √ j
2
2
and
7
1
1
−n= √ i− √ j
2
2
Important Concept
At every point (x0 , y0 ) in the domain of a differentiable function f (x, y), the gradient of
f is normal to the level curve through (x0 , y0 ).
Tangent Line to a Level Curve
fx (x0 , y0 )(x − x0 ) + fy (x0 , y0 )(y − y0 ) = 0
Notice this is the same as point-slope form from elementary algebra.
y − y0 = m(x − x0 )
where
m=−
dy
fx
=
fy
dx
by Theorem 8.
Algebra Rules for Gradients
1.
2.
3.
4.
Sum Rule:
Difference Rule:
Constant Multiple Rule:
Product Rule:
5. Quotient Rule:
∇(f + g) = ∇f + ∇g
∇(f − g) = ∇f − ∇g
∇(kf ) = k∇f
∇(f g) = f ∇g + g∇f
f
g∇f − f ∇g
∇
=
g
g2
Gradients of Functions of n variables
For a differential function f (x1 , x2 , . . . xn ) and a unit vector u = hu1 , u2 , . . . , un i in space, we have
∂f
∂f ∂f
,
,...,
∇f =
∂x1 ∂x2
∂xn
and
n
X ∂f
∂f
∂f
∂f
Du f = ∇f · u =
u1 +
u2 + . . .
un =
ui
∂x1
∂x2
∂xn
∂xi
i=1
The Derivative Along a Path
Let r(t) = x(t)i + y(t)j + z(t)k be a smooth path C and w = f (r(t)) a scalar function along C.
Then
dw
∂w dx ∂w dy ∂w dz
=
+
+
dt
∂x dt
∂y dt
∂z dt
or in vector notation,
d
f (u(t)) = ∇f (r(t)) · r0 (t)
dt
8