# Section 14.3 Partial Derivatives

```Section 14.3 Partial Derivatives
Ruipeng Shen
March 20
1
Basic Conceptions
If f (x, y) is a function of two variables x and y, suppose we let only x vary while keeping y fixed,
say y = b, where b is a constant. Then we are really considering a function of a single variable
x, namely, g(x) = f (x, b). If g has a derivative at a, then we call it the partial derivative of
f with respect to x at (a, b)
fx (a, b) = g 0 (a) = lim
h→0
f (a + h, b) − f (a, b)
.
h
Definition 1. If f is a function of two variables, its partial derivatives are the functions fx
and fy defined by
f (x + h, y) − f (x, y)
;
h→0
h
fx (x, y) = lim
f (x, y + h) − f (x, y)
.
h→0
h
fy (x, y) = lim
There are many alternative notations for partial derivatives of z = f (x, y) as below
∂f
∂
=
f (x, y) =
∂x
∂x
∂f
∂
fy (x, y) = fy =
=
f (x, y) =
∂y
∂y
fx (x, y) = fx =
∂z
= f1 = D 1 f = D x f
∂x
∂z
= f2 = D2 f = Dy f
∂y
Rule for Finding Partial Derivatives
1. To find fx , regard y as a constant and differentiate f (x, y) with respect to x.
2. To find fy , regard x as a constant and differentiate f (x, y) with respect to y.
Example 2. If f (x, y) = x3 + sin(xy), find fx (1, 1) and fy (1, 1).
Solution
Basic calculation shows
fx = 3x2 + y cos(xy);
fy = x cos(xy).
Therefore we have
fx (1, 1) = 3 + cos 1;
fy (1, 1) = cos 1.
1
z
T1
C1
T2
C2
y
(a,b,0)
x
Figure 1: Geometrical Interpretation of Partial Derivatives
2
Interpretations of Partial Derivatives
The partial derivative fx (a, b) and fy (a, b) can be interpreted geometrically as the slopes of the
tangent lines at P (a, b, f (a, b)) to the two traces C1 and C2 of S in the planes y = b and x = a.
Example 3. Suppose that a differentiable function f (x, y) has a maximal value at (1, 3), as
shown in the figure 2. What is the value of the partial derivative fx (1, 3) and fy (1, 3)?
Solution Since the tangent line at the “peak” is always horizontal, we know the slopes fx (1, 3)
and fy (1, 3) must be zero.
z
(1,3,0)
y
x
Figure 2: The maximum has horizontal tangent lines
2
3
More Differentiation
Example 4. If f (x, y) = ex/y , calculate
Solution
∂f
∂f
and
.
∂x
∂y
Using the chain rule we have
∂f
= ex/y ·
∂x
∂f
= ex/y ·
∂y
∂ x
=
∂x y
∂ x
=
∂y y
1 x/y
e ;
y
−x x/y
e .
y2
Example 5. Assume z is defined implicitly as a function of x and y by the equation
x3 + y 3 + z 3 + 6xyz = 1.
Find the partial derivatives ∂z/∂x and ∂z/∂y.
Solution
Applying implicit differentiation with respect to x, we have
3x2 + 3z 2 ·
∂z
∂z
+ 6yz + 6xy ·
=0
∂x
∂x
Solving ∂z/∂x from the identity above, we have
∂z
x2 + 2yz
=− 2
.
∂x
z + 2xy
In the same way we have
y 2 + 2xz
∂z
=− 2
.
∂y
z + 2xy
4
Higher Derivatives
If z = f (x, y) is a function of two variables, then its partial derivatives fx and fy are also
functions of two variables, so we can consider their partial derivatives (fx )x , (fx )y , (fy )x and
(fy )y , which are called the second partial derivatives of f . We use the following notations
∂ ∂f
∂2f
∂2z
(fx )x = fxx = f11 =
=
=
;
2
∂x ∂x
∂x
∂x2
∂ ∂f
∂2f
∂2z
=
=
;
(fx )y = fxy = f12 =
∂y ∂x
∂y∂x
∂y∂x
∂ ∂f
∂2f
∂2z
(fy )x = fyx = f21 =
=
=
;
∂x ∂y
∂x∂y
∂x∂y
∂ ∂f
∂2f
∂2z
(fy )y = fyy = f22 =
=
=
;
∂y ∂y
∂y 2
∂y 2
Example 6. Find the second partial derivatives of f (x, y) = x3 + x2 y 3 − 2y 2 .
3
Solution
We first calculate the first-order partial derivatives
fx = 3x2 + 2xy 3 ;
fy = 3x2 y 2 − 4y.
Therefore we have
∂
fxx =
3x2 + 2xy 3 = 6x + 2y 3 ;
∂x
∂
fyx =
3x2 y 2 − 4y = 6xy 2 ;
∂x
∂
3x2 + 2xy 3 = 6xy 2 ;
∂y
∂
=
3x2 y 2 − 4y = 6x2 y − 4;
∂y
fxy =
fyy
Theorem 7 (Clairaut’s Theorem). Suppose f is defined in a disk D containing the point (a, b).
If the functions fxy and fyx are both continuous on D. then
fxy (a, b) = fyx (a, b).
We can also define derivatives of order 3 or higher. For instance
2 ∂
∂ f
∂3f
fxyy = (fxy )y =
=
.
∂y ∂y∂x
∂y 2 ∂x
Using Clairaut’s Theorem, one can show fxyy = fyxy = fyyx as long as all partial derivatives of
f are continuous.
Example 8. Calculate fxxyz if f (x, y, z) = sin(3x + yz).
Solution
Basic calculation shows
fx = 3 cos(3x + yz);
fxx = −9 sin(3x + yz);
fxxy = −9z cos(3x + yz);
fxxyz = −9 cos(3x + yz) + 9yz sin(3x + yz).
5
Partial Differential Equations
Partial Differential Equations are equations that involve the partial derivatives of an unknown
function. For example, the partial differential equation
∂2u ∂2u
+ 2 =0
∂x2
∂y
is called Laplace’s equation. Solutions to this equation are called harmonic functions.
Example 9. Show that the function u(x, y) = ex sin y is a solution to the Laplace’s equation.
Solution
By a basic calculation we have
ux = ex sin y,
uy = ex cos y;
uxx = ex sin y,
uyy = −ey sin y;
Therefore we have uxx + uyy = 0.
Example 10. Let f be a smooth function. Verify the function u(x, t) = f (x − at) satisfies the
wave equation
2
∂2u
2∂ u
=
a
.
∂t2
∂x2
4
Solution
By a basic calculation we have
ut = −af 0 (x − at)
ux = f 0 (x − at);
utt = a2 f 00 (x − at),
uxx = f 00 (x − at);
Therefore we have utt = a2 f 00 (x − at) = a2 uxx = 0.
5
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