For f(x, y) = −xy² + x² + 2y³, find the partial derivative ∂f/∂y at the point (2, −1).
Multivariable calculus worksheet with answers
1.Easy
f(x, y) = −xy² + x² + 2y³P = (2, −1)
fy(2, −1) =
2.Easy
For f(x, y) = 3x³y³ − 3x³ − y³ + 3xy, find the mixed second partial derivative ∂²f/∂x∂y at (2, 2).
f(x, y) = 3x³y³ − 3x³ − y³ + 3xyP = (2, 2)
fxy(2, 2) =
3.Medium
For f(x, y) = 3 sin(x + y), find the maximum rate of change of f at the point (−1, 0.5), that is |∇f|.
f(x, y) = 3 sin(x + y)P = (−1, 0.5)
|∇f| =
4.Medium
z = −2x³y³ − 4x³ + y² − 3xy with x = t + 1 and y = 2t². Use the chain rule to find dz/dt at t = 1.
z = −2x³y³ − 4x³ + y² − 3xyx = t + 1y = 2t²t = 1
dz/dt =
5.Hard
The marked points are the critical points of f(x, y) = x³ + y³ + 3xy. Click the saddle point.
f(x, y) = x³ + y³ + 3xy
- 1A (−1, −1)
- 2B (0, 0)
6.Hard
Find the area I = ∬R 1 dA of the region R between the parabola y = x² and the line y = 3x.
y = x²y = 3xf = 1
I =
Answer key
- 1. fy(2, −1) = 10
- 2. fxy(2, 2) = 435
- 3. |∇f| = 3.723
- 4. dz/dt = −1022
- 5. B (0, 0)
- 6. I = 4.5
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