Math 3220-1
HW 4.3
Solutions
Exercises for Section 4.3: Partial Derivatives
1. Find the partial derivatives of the following functions:
(a) f (x, y, z) = xy .
(b) f (x, y, z) = (xy , z).
(c) f (x, y, z) = sin(x sin y).
(d) f (x, y, z) = sin (x sin(y sin z)).
(e) f (x, y, z) = x(y ) . (That is “x to the (y z ) power.”)
z
(f) f (x, y, z) = xy+z .
(g) f (x, y, z) = (x + y)z .
(h) f (x, y) = sin(xy).
(i) f (x, y) = (sin(xy))cos 3 .
(j) f (x, y) = (sin(xy), sin(x sin y), xy ).
Solution. Here are the non-zero partial derivatives.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
∂f
= yxy − 1,
∂x
∂f
= xy ln x.
∂y
∂f 1
∂f 2
∂f 1
= yxy − 1,
= xy ln x,
= 1.
∂x
∂y
∂z
∂f
∂f
= sin y cos(x sin y),
= x cos(x sin y) cos y.
∂x
∂y
∂f
= sin(y sin z) cos(x sin(y sin z)),
∂x
∂f
= x cos(x sin(y sin z)) cos(y sin z) sin z,
∂y
∂f
= cos(x sin(y sin z))x cos(y sin z)y cos z.
∂z
z
z
z
∂f
∂f
∂f
= (y z ) x(y −1) ,
= x(y ) ln xzy z−1 ,
= x(y ) ln xy z ln y.
∂x
∂y
∂z
∂f
∂f
∂f
= (y + z)x(y+z−1) ,
=
= xy+z ln x.
∂x
∂x
∂z
∂f
∂f
∂f
=
= z(x + y)z−1 ,
= (x + y)z ln(x + y).
∂x
∂y
∂z
∂f
∂f
= y cos(xy),
= x cos(xy).
∂x
∂y
∂f
∂f
= cos 3 (sin(xy))cos 3−1 y cos(xy),
= cos 3 (sin(xy))cos 3−1 x cos(xy).
∂x
∂y
∂f 1
∂f 1
= y cos(xy),
= x cos(xy),
∂x
∂y
∂f 2
∂f 2
= sin y cos(x sin y),
= x cos(x sin y) cos y,
∂x
∂y
∂f 3
∂f 3
= yxy−1 ,
= xy ln x.
∂x
∂y
2. Find the partial derivatives of the following functions (where g : R −→ R is continuous).
Z x+y
(a) f (x, y) =
g.
a
Z xy
(b) f (x, y) =
g.
a
Z
(c) f (x, y, z) =
sin(x sin(y sin z))
g.
xy
∂f
∂f
=
= g(x + y).
∂x
∂y
∂f
∂f
= yg(xy),
= xg(xy).
(b)
∂x
∂y
∂f
= g (sin (x sin(y sin z))) cos (x sin(y sin z)) sin(y sin z) − yg(xy),
(c)
∂x
∂f
= g (sin (x sin(y sin z))) cos (x sin(y sin z)) x cos(y sin z) sin z − xg(xy),
∂y
∂f
= g (sin (x sin(y sin z))) cos (x sin(y sin z)) x cos(y sin z)y cos z.
∂z
Solution. (a)
3. If
f (x, y) = xx
find
xx
y
+ (log x)(arctan(arctan(arctan(sin(cos xy) − log(x + y))))),
∂f
(1, y).
∂y
Solution. Define g(y) = f (1, y) = 1. Then
∂f
(1, y) = g ′ (y) = 0.
∂y
4. Find the partial derivatives of f in terms of the derivatives of g and h if
(a) f (x, y) = g(x)h(y).
(b) f (x, y) = g(x)h(y) .
(c) f (x, y) = g(x).
(d) f (x, y) = g(y).
(e) f (x, y) = g(x + y).
Solution. (a)
∂f
∂x
∂f
(c)
∂x
∂f
(d)
∂x
∂f
(e)
∂x
(b)
∂f
= g ′ (x)h(y),
∂x
= h(y)g(x)h(y)−1 g ′ (x),
= g ′ (x),
= 0,
=
∂f
= 0.
∂y
∂f
= g ′ (y).
∂y
∂f
= g ′ (x + y).
∂y
∂f
= g(x)h′ (y).
∂y
∂f
= g(x)h(y) ln(g(x))h′ (y).
∂y
5. Let g1 , g2 : R2 −→ R be continuous. Define f : R2 −→ R by
Z x
Z y
f (x, y) =
g1 (t, 0) dt +
g2 (x, t) dt.
0
(a) Show that
0
∂f
(x, y) = g2 (x, y).
∂y
∂f
(x, y) = g1 (x, y)?
∂x
∂f
∂f
(x, y) = x, and
(x, y) = y. Find one such that
(c) Find a function f : R2 −→ R so that
∂x
∂y
∂f
∂f
(x, y) = y, and
(x, y) = x.
∂x
∂y
(b) How should f be defined so that
Solution. (a) This follows from the First Fundamental Theorem of Calculus.
Rx
(b) Define f (x, y) = 0 g1 (t, y) dt.
(c) Define f1 (x, y) = 21 (x2 + y 2 ). Define f2 (x, y) = xy. These satisfy the required conditions.
6. If f : R2 −→ R and
that f is constant.
∂f
∂f
∂f
= 0, show that f is independent of the second variable. If
=
= 0, show
∂y
∂x
∂y
∂f
(x, y) = 0. Now, let y1 , y2 ∈ R. By the
∂y
Mean Value Theorem from Analysis I, g(y1 ) = g(y2 ). Therefore, f (x, y1 ) = f (x, y2 ).
∂f
∂f
=
= 0, then apply this argument twice to conclude f is constant.
If
∂x
∂y
Proof. Fix x ∈ R, and define g(y) = f (x, y). Then g ′ (y) =
7. Let
A = (x, y) ∈ R2 : x < 0, or x ≥ 0 and y 6= 0 .
∂f
∂f
=
= 0, show that f is constant. Hint: You cannot use the previous
∂x
∂y
problem because the domain of f is not R2 . However, note that any two points in A can be
connected by a sequence of lines, each parallel to one of the axes.
∂f
(b) Find a function f : A −→ R such that
= 0, but f is not independent of the second variable.
∂y
Contrast this with the result in Problem 6, above.
(a) If f : A −→ R and
Solution. (a) We could do this by considering different cases. But I will demonstrate how to show
this without resorting to cases.
Proof. Let (x1 , y1 ) and (x2 , y2 ) be two points in A. We need to show that f (x1 , y1 ) = f (x2 , y2 ).
Consider the line segment from the point (x1 , y1 ) to the point (−|x1 | − 1, y1 ). The function f will
be constant along this line segment. This can be shown using the Mean Value Theorem, as in the
problem above. Next consider the line segment between (−|x1 | − 1, y1 ) and (−|x1 | − 1, y2 ). Again,
f is constant along this line segment. Last, consider the line segment between (−|x1 | − 1, y2 ) and
(x2 , y2 ). Once again, f is constant along this line segment. Therefore f (x1 , y1 ) = f (x2 , y2 ), hence
f is constant on A.
(b) Consider the function f : A −→ R via:

 0
x2
f (x, y) =
 2
−x
if
if
if
x < 0,
x > 0, y > 0,
x > 0, y < 0.
Note that f is not independent of the second variable because f (1, 1) 6= f (1, −1). Other examples
are certainly possible.
8. Define f : R2 −→ R by
(a) Show that
x2 − y 2
, if (x, y) 6= 0,
f (x, y) =
x2 + y 2

0,
if (x, y) = 0.


xy
∂f
∂f
(x, 0) = x for all x, and
(0, y) = −y for all y.
∂y
∂x
∂2f
∂2f
(0, 0) 6=
(0, 0).
∂y∂x
∂x∂y
(c) What is the significance of this example?
(b) Show that
Solution. (a) As long as (x, y) 6= 0, this is a simple calculation using familiar techniques of differentiation. Consider now (x, y) = 0.
∂f
(0, 0) =
∂x
f (0 + t, 0) − f (0, 0)
t
0−0
= 0.
= lim
t→0
t
A similar calculation shows that
lim
t→0
∂f
(0, 0) = 0.
∂y
∂2f
∂2f
(0, 0) = −1 while
(0, 0) = 1.
∂y∂x
∂x∂y
(c) Because the mixed partial derivatives are unequal, we may conclude that the partial derivatives fail
to be continuous on any open neighborhood of the origin. Moreover, this function is differentiable
at the origin, with df0 = 0. However, it fails to be continuously differentiable.
(b)
9. Define f : R −→ R by
f (x) =
−2
e−x ,
0,
if x 6= 0,
if x = 0.
Show that f is a C ∞ function, and f (n) (0) = 0 for all n. Hint: The limit
e−h
f (0) = lim
h→0
h
−2
′
= lim
h→0
h
e
h−2
can be evaluated by l’Hôpital’s rule. It is easy enough to find f ′ (x) for x 6= 0, and f ′′ (0) can then be
found by another application of l’Hôpital’s rule.
Note that this function is equal up to nth order to 0 for all n (See Problem ??.??).
Proof. We will prove this by induction on n. Our induction hypothesis is that
−2
pn (1/x)e−x , if x 6= 0,
(n)
f (x) =
0,
if x = 0,
where pn (z) is a polynomial in z.
The base case is n = 0, for which this is true. Now suppose this is true for the first n derivatives, and
compute f (n+1) (x). If x 6= 0, then we can compute the derivative using Calculus techniques:
f (n+1) (x) = pn (1/x)(−2x−3 )e−x
−2
+ p′n (1/x)(−x−2 )e−x
−2
−2
= pn+1 (1/x)e−x ,
where pn+1 (z) = −2z 3 pn (z) − z 2 p′n (z) is a polynomial in z.
If x = 0 then
f (n+1) (0) =
=
=
pn (1/h)e−h
h→0
h
1
p
(1/h)
n
lim h h−2
h→0
e
µpn (µ)
.
lim
µ→∞
eµ2
−2
lim
This last limit is clearly 0, for polynomials grow much slower than the exponential.
f (n+1) (0) = 0. Hence, by induction, the claim holds.
(1)
(2)
(3)
Therefore,