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Chapter 14 Partial Derivatives
14.1 Functions of Several Variables
A real-valued function of two variables, f , is a rule in terms of x and y, denoted
by
z = f(x, y)
where x and y are real numbers.
Ex 1. Let f be the function defined by
f(x, y) = 4xy + 6 sin x− 2y.
Compute f(0, 1) and f(π, 2).
Ex 2. Let g be the function defined by
g(x, y) = ex − 1 x− y .
Compute g(1, 0), g(1, e) and g(e, 1).
1
The domain of a function f is the set of points (x, y) such that z = f(x, y) is a real
number. Unless specified, the domain of a function f will be taken to be the largest
possible set for which the rule defining f is meaningful.
Ex 3. Find and sketch the domain of each of the following functions.
1. f(x, y) = 4xy + 6 sin x− 2y
2. f(x, y) = ex − 1 x− y
3. f(x, y) = √
1− x2 − y2
4. f(x, y) = ln(x2 − y)
2
14.2 Limits and Continuity
Ex 4. [Plugging in] Evaluate
1. lim (x,y)→(1,1)
3x 2+y2 cos(
π
2x+ y )
2. lim (x,y)→(2,−1)
ln
(
1 + x+ 2y
3y2 − x
)
3. lim (x,y)→(−1,1)
arctan xy
2x2 + y2
3
Ex 5. [Factoring] Evaluate
1. lim (x,y)→(2,1)
x2 − 4y2 x2 − 2xy
2. lim (x,y)→(1,−1)
x+ y
x− √
xy + 2y2
4
3. lim (x,y)→(−1,0)
arctan y2 − xy y √ 3 + xy
4. lim (x,y)→(1,2)
2x3 + 6xy − x2y − 3y2 2x4 + 2xy2 − x3y − y3
5
Limits along curves
Ex 6. Compute the following limits.
1. lim (x,y)→(0,0) on y = x
x2
x2 + y2
2. lim (x,y)→(0,0) on x = y2
2x2y
x+ y2
3. lim (x,y)→(0,0) on y = −x
x2y3
x6 + y6
6
Ex 7. Let f(x, y) = xy
x2 + y2 , (x, y) 6= (0, 0).
Find the limit of f(x, y) when (x, y) → (0, 0) on the following curves.
1. C1 : X-axis
2. C2 : y = x
3. C3 : y = 2x
7
Consider the point (x0, y0). If C1 and C2 are two curves passing through (x0, y0) and
lim (x,y)→(x0,y0)
on C1
f(x, y) 6= lim (x,y)→(x0,y0)
on C2
f(x, y)
then lim (x,y)→(x0,y0)
f(x, y) does not exist.
Ex 8. Show that the following limits do not exists.
1. lim (x,y)→(0,0)
x
x2 + y2
8
2. lim (x,y)→(0,0)
x4 + y2
x4 − y2
3. lim (x,y)→(0,0)
xy
x2 + y2
9
4. lim (x,y)→(0,0)
x2y
x4 + y2
5. lim (x,y)→(0,0)
sin(x2)
x2 + y2
10
We say that z = f(x, y) is continuous at the points (x0, y0) if
lim (x,y)→(x0,y0)
f(x, y) = f(x0, y0).
Ex 9. Determine the set of points at which the function is continuous.
1. f(x, y) = y
x2 + y2
2. f(x, y) = ln(x2 − y + 4)
Do Exercises. 14.1 9–10 (a) and (b), 11, 13–19 (odd)
Do Exercises. 14.2 5–17, 29–33 (odd)
11
14.3 Partial Derivatives
Let z = f(x, y).
The partial derivative of f with respect to x is
∂f
∂x = lim
h→0
f(x+ h, y)− f(x, y) h
if limit exists
and the partial derivative of f with respect to y is
∂f
∂y = lim
k→0
f(x, y + k)− f(x, y) k
if limit exists
Note that ∂f ∂x
is the derivative of f with respect to x when y is a constant. Similarly, ∂f
∂y is the derivative of f with respect to y when x is a constant.
Thus we could apply the derivative formulas for one variable in CAL I for finding the
partial derivatives of f .
We write ∂f
∂x = fx = f1 = D1f
and ∂f
∂y = fy = f2 = D2f.
E.g. f(x, y) = y2 sin x f(x, y) = √ x ln y
12
Ex 10. Find the partial derivatives of
1. f(x, y) = x3 √ y + ey + arctan x
2. f(x, y) = e(x sin y)
13
3. f(x, y) = ( 3 √ x+ y3)4
4. f(x, y) = x2 tan(x+ 2y)
14
5. f(x, y) = x+ ln y √ y
6. f(x, y) = ysinx
15
7. f(x, y, z) = z(sec y2)(ln x) + (cos z)y
16
Higher-Order Partial Derivatives
fxx = ∂2z
∂x2
fx = ∂z
∂x
✉✉✉✉✉✉✉✉✉
■■ ■■
■■ ■■
■
fxy = ∂2z
∂y∂x
z = f(x, y)
✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟✟
✻✻ ✻✻
✻✻ ✻✻
✻✻ ✻✻
✻✻ ✻✻
✻✻ ✻✻
fyx = ∂2z
∂x∂y
fy = ∂z
∂y
✈✈✈✈✈✈✈✈✈
❍❍ ❍❍
❍❍ ❍❍
❍
fyy = ∂2z
∂y2
The last four vertices of the above diagrams are called the second order partial deriva-
tives of f .
Ex 11. Find the second order partial derivatives of f .
1. f(x, y) = x2y3 + √ x ln y
17
2. f(x, y) = x arctan y − y4e2x
3. f(x, y) = ln(3x+ y2)
18
Ex 12. Let f(x, y) = (sin y)(tan x) + (x2 + 3)y. Find fxy and fyx.
Do Exercises. 14.3 15–27 (odd), 31–37 (odd), 41, 43, 51–57 (odd), 63–69 (odd)
19
Interpretations of Partial Derivatives
Partial derivatives can also be interpreted as rates of change. If z = f(x, y), then ∂f
∂x = the rate of change of z with respect to x when y is fixed, and
∂f
∂y = the rate of change of z with respect to y when x is fixed.
Ex 13. Let z = f(x, y) = 4− x2 − 2y2. At the point (1, 1, 1), we have
∂z
∂x = −2x and ∂z
∂x
∣
∣
∣
∣
(1,1)
= −2
is the slope of the tangent line to the curve C1 which is the intersection of the plane
y = 1 and z = f(x, y) as shown.
Similarly, ∂z
∂y = −4y ∂z
∂x
∣
∣
∣
∣
(1,1)
= −4
is the slope of the tangent line to the curve C2 which is the intersection of the plane
x = 1 and z = f(x, y) as shown.
20
In general, if we have a surface z = f(x, y), at the point (a, b) we have
fx(a, b) = the slope of the tangent line to the curve which is the intersection of the
plane y = b and z = f(x, y) and
fy(a, b) = the slope of the tangent line to the curve which is the intersection of the
plane x = a and z = f(x, y)
Ex 14. Find the slope of the line tangent to the curve in which
1. the Y Z plane intersects the surface z = arctan(ye2x) at the point (0, 1, π/4)
2. the plane y = −1 intersects the surface 4x2− 3y2+ z = 5 at the point (1,−1, 4)
21
14.4 Tangent Planes and Linear Approximations
Tangent Planes
Suppose f has continuous partial derivatives. An equation of the tangent plane to
the surface z = f(x, y) at the point P (x0, y0, z0) is
z − z0 = fx(x0, y0)(x− x0) + fy(x0, y0)(y − y0).
22
Ex 15. Find an equation of a plane tangent to
1. z = x2 + y2 at the point (2,−1, 5)
2. z = x− y + arctan(xy) at the point (1, 1, π/4)
Do Exercises. 14.4 1–6 (odd)
23
Linear Approximations
From tangent plane to z = f(x, y) at (a, b, c)
z − c = fx(a, b)(x− a) + fy(a, b)(y − b),
we may write
dz = fx dx+ fy dy,
called the total differential of f . For ∆x and ∆y are small, we have
∆z = f(a+∆x, b+∆y)− f(a, b)
≈ fx(a, b)∆x+ fy(a, b)∆y.
That is,
f(a+∆x, b+∆y) ≈ f(a, b) + fx(a, b)∆x+ fy(a, b)∆y.
24
Ex 16. 1. If z = f(x, y) = x2 + 3xy − y2, find the differential dz.
2. If x changes from 2 to 2.05 and y changes from 3 to 2.96, compare the values
dz and ∆z.
25
Ex 17. Approximate √
(3.05)2 − (1.97)3.
26
Ex 18. Find the linear approximation of the function
f(x, y) = y2
(1 + sin x)y
at (0, 2) and use it to approximate f(0.03, 1.98).
Do Exercises. 14.4 11–15 (odd), 25–31 (odd)
27
14.5 The Chain Rule
In CAL I, for y = f(x) and x = g(t), we have
dy
dt =
dy
dx
dx
dt .
y dy dx
x dx dt
t
z ∂z ∂x
⑧⑧ ⑧⑧ ⑧⑧ ⑧⑧
∂z ∂y
❄❄ ❄❄
❄❄ ❄❄
x dx dt
y dy dt
t t
The Chain Rule (Case 1) Suppose that z = f(x, y) is a differentiable function of x
and y, where x = g(t) and y = h(t) are both differentiable functions of t. Then
dz
dt =
∂z
∂x
dx
dt +
∂z
∂y
dy
dt .
Ex 19. 1. If z = x2 − 2xy + xy3, x = t3 + 1 and y = 1/t, compute dz dt
at t = 1.
2. If z = x2 + wey + sin(xw), x = 2t, y = t2 and w = sec t, compute dz
dt .
28