For f(x, y) = xy² + 2x³ + y³ − 4xy, find the partial derivative ∂f/∂x at the point (−2, −1).
Multivariable calculus worksheet with answers
1.Easy
f(x, y) = xy² + 2x³ + y³ − 4xyP = (−2, −1)
fx(−2, −1) =
2.Easy
For f(x, y) = ln(x² + 4y²), find the mixed second partial derivative ∂²f/∂x∂y at (−1, 1).
f(x, y) = ln(x² + 4y²)P = (−1, 1)
fxy(−1, 1) =
3.Medium
For f(x, y) = 2x²e−y, find the direction of steepest ascent at (1, 0.5) as the angle θ from the positive x-axis, in degrees between −180° and 180°.
f(x, y) = 2x²e−yP = (1, 0.5)
θ = °
4.Medium
z = −4xy − 2x³ + 2y² with x = −3t + 1 and y = t². Use the chain rule to find dz/dt at t = −1.
z = −4xy − 2x³ + 2y²x = −3t + 1y = t²t = −1
dz/dt =
5.Hard
f(x, y) = −x³ + 3x − y² − 6y has a critical point at (1, −3). Compute the discriminant D = fxxfyy − fxy² there.
f(x, y) = −x³ + 3x − y² − 6yP = (1, −3)
D =
6.Hard
The marked points are the critical points of f(x, y) = x⁴ + y⁴ − 16xy. Click the saddle point.
f(x, y) = x⁴ + y⁴ − 16xy
- 1A (−2, −2)
- 2B (2, 2)
- 3C (0, 0)
Answer key
- 1. fx(−2, −1) = 29
- 2. fxy(−1, 1) = 0.64
- 3. θ = −26.57 °
- 4. dz/dt = 324
- 5. D = 12
- 6. C (0, 0)
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