University of Aarhus
Department of Mathematical Sciences
Calculus Quiz
week 36.2
Legend: A 4 indicates a correct response; a 8, indicates an incorrect response, in this case, the correct answer is marked with a l. Correct and follow
a 4,l for a solution. Return through n.
Fall 2004
2
Partial derivatives
1. If f (x, y) = x2 − 2y 2 + 2x + 5, then
fx = 2x − 4y + 2
fy = 2x − 4y + 2
fx = 2x + 2
fy = x2 − 4y + 2x
2. The partial derivatives of the function f (x, y) = ln(xy + 1) are
y
x
fx =
, fy =
xy + 1
xy + 1
Yes
No
3. The function f (x, y) = sin(y − x2 ) has
fyy (x, y) = −f (x, y)
Yes
No
3
4. For any sufficiently nice function f (x, y, z),
fxy = fyx
Yes
No
5. If f (x, y) is a nice function with fxx = 0, then
fxyxy = 0
Yes
No
6. If f (x, y) is a function in R2 with partial derivaties, then the function
g(x, y, z) = f (x, y) · z
has partial derivatives in R3 .
Yes
No
Solutions to Quizzes
4
Solutions to Quizzes
Solution to Quiz:
f (x, y) = x2 − 2y 2 + 2x + 5 gives
fx =
d 2
(x − 2y 2 + 2x + 5) = 2x + 2
dx
Solutions to Quizzes
5
Solution to Quiz:
Yes, Given f (x, y) = ln(xy + 1).
By the usual chain rule for differentiation
1
d
y
d
ln(xy + 1) =
(xy + 1) =
dx
xy + 1 dx
xy + 1
The other by symmetry.
fx =
Solutions to Quizzes
6
Solution to Quiz:
Yes, f (x, y) = sin(y − x2 ) gives:
fyy =
d2
d
cos(y − x2 ) = − sin(y − x2 ) = −f (x, y)
sin(y − x2 ) =
dy 2
dy
Solutions to Quizzes
Solution to Quiz:
Yes, This is Clairault’s theorem [S] 11.3.
7
Solutions to Quizzes
8
Solution to Quiz:
Yes, If fxx = 0, then by Clairault’s theorem [S] 11.3
fxyxy = fyyxx = (fxx )yy = 0yy = 0
Solutions to Quizzes
Solution to Quiz:
Yes, by primitive rules for partial derivatives. See [S] 11.3.
9