Master multivariable calculus with 10 comprehensive problems on partial derivatives, gradients, and chain rule.
Find ∂f/∂x and ∂f/∂y. Then compute ∇f(1,2).
1 Partial derivative with respect to x (treat y as constant):
∂f/∂x = ∂/∂x[3x²y] + ∂/∂x[2xy³] - ∂/∂x[5x] + ∂/∂x[7y]
= 6xy + 2y³ - 5 + 0
∂f/∂x = 6xy + 2y³ - 5
2 Partial derivative with respect to y (treat x as constant):
∂f/∂y = ∂/∂y[3x²y] + ∂/∂y[2xy³] - ∂/∂y[5x] + ∂/∂y[7y]
= 3x² + 6xy² - 0 + 7
∂f/∂y = 3x² + 6xy² + 7
3 Gradient vector:
∇f = ⟨6xy + 2y³ - 5, 3x² + 6xy² + 7⟩
4 Evaluate at (1,2):
∂f/∂x(1,2) = 6(1)(2) + 2(2)³ - 5 = 12 + 16 - 5 = 23
∂f/∂y(1,2) = 3(1)² + 6(1)(2)² + 7 = 3 + 24 + 7 = 34
∇f(1,2) = ⟨23, 34⟩
Find all second partial derivatives: fxx, fyy, fxy, fyx. Verify Clairaut's theorem.
1 First partial derivatives:
f_x = ∂/∂x[x³e^y + y·sin(x)] = 3x²e^y + y·cos(x)
f_y = ∂/∂y[x³e^y + y·sin(x)] = x³e^y + sin(x)
2 Second partial derivatives:
f_xx = ∂/∂x[f_x] = ∂/∂x[3x²e^y + y·cos(x)] = 6xe^y - y·sin(x)
f_yy = ∂/∂y[f_y] = ∂/∂y[x³e^y + sin(x)] = x³e^y
3 Mixed partials:
f_xy = ∂/∂y[f_x] = ∂/∂y[3x²e^y + y·cos(x)] = 3x²e^y + cos(x)
f_yx = ∂/∂x[f_y] = ∂/∂x[x³e^y + sin(x)] = 3x²e^y + cos(x)
4 Verify Clairaut's theorem:
f_xy = 3x²e^y + cos(x)
f_yx = 3x²e^y + cos(x)
∴ f_xy = f_yx ✓
Find dz/dt using the chain rule. Then find dz/dt at t = π/4.
1 Chain rule formula for one independent variable:
dz/dt = (∂z/∂x)(dx/dt) + (∂z/∂y)(dy/dt)
2 Compute partial derivatives:
∂z/∂x = 2xy
∂z/∂y = x² - 2y
3 Compute ordinary derivatives:
dx/dt = cos(t)
dy/dt = -sin(t)
4 Apply chain rule:
dz/dt = (2xy)(cos(t)) + (x² - 2y)(-sin(t))
= 2xy·cos(t) - x²·sin(t) + 2y·sin(t)
5 Substitute x = sin(t), y = cos(t):
= 2·sin(t)·cos(t)·cos(t) - sin²(t)·sin(t) + 2·cos(t)·sin(t)
= 2sin(t)cos²(t) - sin³(t) + 2sin(t)cos(t)
6 Evaluate at t = π/4:
sin(π/4) = √2/2, cos(π/4) = √2/2
dz/dt = 2(√2/2)(1/2) - (√2/2)³ + 2(√2/2)(√2/2)
= √2/2 - (√2/8) + 1 = (4√2/8 - √2/8) + 1 = (3√2/8) + 1
≈ 0.53033 + 1 = 1.53033
Given z is defined implicitly as a function of x and y, find ∂z/∂x and ∂z/∂y at point (1,0,0).
1 Differentiate implicitly with respect to x (treat y as constant, z as z(x,y)):
∂/∂x[x²] + ∂/∂x[y²] + ∂/∂x[z²] + ∂/∂x[xyz] = ∂/∂x[1]
2x + 0 + 2z·(∂z/∂x) + yz + xy·(∂z/∂x) = 0
(2z + xy)·(∂z/∂x) = -2x - yz
∂z/∂x = -(2x + yz)/(2z + xy)
2 Differentiate implicitly with respect to y:
∂/∂y[x²] + ∂/∂y[y²] + ∂/∂y[z²] + ∂/∂y[xyz] = ∂/∂y[1]
0 + 2y + 2z·(∂z/∂y) + xz + xy·(∂z/∂y) = 0
(2z + xy)·(∂z/∂y) = -2y - xz
∂z/∂y = -(2y + xz)/(2z + xy)
3 Evaluate at (1,0,0):
∂z/∂x = -(2(1) + (0)(0))/(2(0) + (1)(0)) = -2/0 (undefined)
∂z/∂y = -(2(0) + (1)(0))/(2(0) + (1)(0)) = 0/0 (indeterminate)
4 Check the point satisfies equation:
1² + 0² + 0² + (1)(0)(0) = 1 ✓
Find the directional derivative of f at point P(1,2) in the direction of vector v = ⟨3,-4⟩. Also find the direction of steepest ascent at P.
1 Compute gradient at P:
∇f = ⟨∂f/∂x, ∂f/∂y⟩ = ⟨2x - 3y, -3x + 8y⟩
∇f(1,2) = ⟨2(1)-3(2), -3(1)+8(2)⟩ = ⟨2-6, -3+16⟩ = ⟨-4, 13⟩
2 Normalize direction vector v:
||v|| = √(3² + (-4)²) = √(9+16) = √25 = 5
Unit vector u = v/||v|| = ⟨3/5, -4/5⟩
3 Directional derivative formula:
D_u f = ∇f · u = ⟨-4, 13⟩ · ⟨3/5, -4/5⟩
= (-4)(3/5) + (13)(-4/5) = (-12/5) + (-52/5) = -64/5 = -12.8
4 Direction of steepest ascent:
Steepest ascent is in direction of gradient: ∇f(1,2) = ⟨-4, 13⟩
Unit vector in this direction: ∇f/||∇f|| = ⟨-4, 13⟩/√(16+169) = ⟨-4, 13⟩/√185
5 Maximum rate of increase:
||∇f(1,2)|| = √(16+169) = √185 ≈ 13.601
| Concept | Formula | Interpretation |
|---|---|---|
| Partial Derivative | ∂f/∂x = limh→0[f(x+h,y)-f(x,y)]/h | Slope in x-direction |
| Gradient | ∇f = ⟨f_x, f_y⟩ | Direction of steepest ascent |
| Directional Derivative | D_u f = ∇f · u | Rate of change in direction u |
| Chain Rule (1 var) | dz/dt = f_x·dx/dt + f_y·dy/dt | Total derivative |
| Laplacian | ∇²f = f_xx + f_yy | Sum of second partials |