For f(x, y) = −2x²e2y, find the mixed second partial derivative ∂²f/∂x∂y at (−1, 2).
Multivariable calculus worksheet with answers
1.Easy
f(x, y) = −2x²e2yP = (−1, 2)
fxy(−1, 2) =
2.Easy
For f(x, y) = −2x²e3y, find the partial derivative ∂f/∂y at the point (0.5, 1).
f(x, y) = −2x²e3yP = (0.5, 1)
fy(0.5, 1) =
3.Medium
For f(x, y) = ln(x² + 2y²), find the maximum rate of change of f at the point (0.5, 0.5), that is |∇f|.
f(x, y) = ln(x² + 2y²)P = (0.5, 0.5)
|∇f| =
4.Medium
For f(x, y) = xe2y, find the maximum rate of change of f at the point (0.5, −1), that is |∇f|.
f(x, y) = xe2yP = (0.5, −1)
|∇f| =
5.Hard
Use polar coordinates to evaluate I = ∬R (x² + y²) dA, where R is the upper half of the disc x² + y² ≤ 2.25 (y ≥ 0).
f = x² + y²0 ≤ r ≤ 1.50 ≤ θ ≤ π
I =
6.Hard
f(x, y) = x³ − 12x + y² − 6y has a critical point at (−2, 3). Compute the discriminant D = fxxfyy − fxy² there.
f(x, y) = x³ − 12x + y² − 6yP = (−2, 3)
D =
Answer key
- 1. fxy(−1, 2) = 436.8
- 2. fy(0.5, 1) = −30.13
- 3. |∇f| = 2.981
- 4. |∇f| = 0.1914
- 5. I = 3.976
- 6. D = −24
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