For f(x, y) = 2 ln(x² + 3y²), find the mixed second partial derivative ∂²f/∂x∂y at (1, −1).
Multivariable calculus worksheet with answers
1.Easy
f(x, y) = 2 ln(x² + 3y²)P = (1, −1)
fxy(1, −1) =
2.Easy
Evaluate I = ∫14 ∫24 (−3x²y + 5) dy dx over the rectangle 1 ≤ x ≤ 4, 2 ≤ y ≤ 4.
f = −3x²y + 51 ≤ x ≤ 42 ≤ y ≤ 4
I =
3.Medium
z = 2x²ey with x = −3t − 2 and y = −2t² + 1. Use the chain rule to find dz/dt at t = 1.
z = 2x²eyx = −3t − 2y = −2t² + 1t = 1
dz/dt =
4.Medium
z = 3x³y² + 3x³ − 5y³ − 2xy with x = t − 1 and y = t² − 1. Use the chain rule to find dz/dt at t = −1.
z = 3x³y² + 3x³ − 5y³ − 2xyx = t − 1y = t² − 1t = −1
dz/dt =
5.Hard
Find the critical points of f(x, y) = x² − 3xy + 3y² − 5x + 9y − 3, classify them, and give the value fmin of f at its local minimum.
f(x, y) = x² − 3xy + 3y² − 5x + 9y − 3
fmin =
6.Hard
The marked points are the critical points of f(x, y) = x⁴ + y⁴ − 4xy. Click the saddle point.
f(x, y) = x⁴ + y⁴ − 4xy
- 1A (1, 1)
- 2B (0, 0)
- 3C (−1, −1)
Answer key
- 1. fxy(1, −1) = 1.5
- 2. I = −348
- 3. dz/dt = −51.5
- 4. dz/dt = 28
- 5. fmin = −10
- 6. B (0, 0)
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