For f(x, y) = −y·exy, find the partial derivative ∂f/∂x at the point (−1, 1).
Multivariable calculus worksheet with answers
1.Easy
f(x, y) = −y·exyP = (−1, 1)
fx(−1, 1) =
2.Easy
For f(x, y) = 2y·exy, find the mixed second partial derivative ∂²f/∂x∂y at (0.5, 0.5).
f(x, y) = 2y·exyP = (0.5, 0.5)
fxy(0.5, 0.5) =
3.Medium
For f(x, y) = 2 ln(x² + y²), find the maximum rate of change of f at the point (−1, 0.5), that is |∇f|.
f(x, y) = 2 ln(x² + y²)P = (−1, 0.5)
|∇f| =
4.Medium
z = −3xy³ + 5x² + 2y³ − 4xy with x = 3t + 1 and y = 2t² + 2. Use the chain rule to find dz/dt at t = 0.
z = −3xy³ + 5x² + 2y³ − 4xyx = 3t + 1y = 2t² + 2t = 0
dz/dt =
5.Hard
The integral I = ∫00.8 ∫y0.8 ex² dx dy cannot be done in this order. Reverse the order of integration and evaluate it.
f = ex²0 ≤ y ≤ x ≤ 0.8
I =
6.Hard
Evaluate I = ∬R y dA, where R is the region between the parabola y = x² and the line y = 4x.
y = x²y = 4xf = y
I =
Answer key
- 1. fx(−1, 1) = −0.3679
- 2. fxy(0.5, 0.5) = 2.889
- 3. |∇f| = 3.578
- 4. dz/dt = −66
- 5. I = 0.4482
- 6. I = 68.27
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