DerivativeCalculus.com

Gradient & Directional Derivatives Mastery

Master multivariable calculus with 15 comprehensive problems. Learn gradient vectors, directional derivatives, and optimization techniques.

📚 Core Concepts & Formulas
Gradient Vector (∇f)
∇f(x,y) = ⟨fₓ(x,y), fᵧ(x,y)⟩
∇f(x,y,z) = ⟨fₓ, fᵧ, f_z⟩

Points in direction of steepest ascent

Directional Derivative
Dᵤf(x,y) = ∇f(x,y) · u
where u is a unit vector

Rate of change in direction u

Unit Vector Formula
u = ⟨a,b⟩ / √(a² + b²)
|∇f| = √(fₓ² + fᵧ²)

Magnitude of gradient = max rate of change

Key Insight: The directional derivative is maximum in the direction of ∇f, and its maximum value is |∇f|.

Problem 1: Basic Gradient Computation Beginner
f(x,y) = 2x²y + sin(xy)
Find the gradient vector ∇f at the point (1, π).
1 Compute partial derivatives:
fₓ = ∂/∂x [2x²y + sin(xy)] = 4xy + y·cos(xy)
fᵧ = ∂/∂y [2x²y + sin(xy)] = 2x² + x·cos(xy)
2 Evaluate at (1, π):
fₓ(1, π) = 4(1)(π) + π·cos(π) = 4π - π = 3π
fᵧ(1, π) = 2(1)² + 1·cos(π) = 2 - 1 = 1
∇f(1, π) = ⟨3π, 1⟩
Problem 2: Directional Derivative Intermediate
f(x,y) = x³ + y² - 3xy
Find the directional derivative at (1,2) in the direction of vector v = ⟨3,4⟩.
1 Compute gradient:
∇f = ⟨3x² - 3y, 2y - 3x⟩
∇f(1,2) = ⟨3(1)² - 3(2), 2(2) - 3(1)⟩ = ⟨-3, 1⟩
2 Find unit vector in direction v:
|v| = √(3² + 4²) = √25 = 5
u = v/|v| = ⟨3/5, 4/5⟩
3 Compute directional derivative:
Dᵤf(1,2) = ∇f(1,2) · u = ⟨-3, 1⟩ · ⟨3/5, 4/5⟩
= (-3)(3/5) + (1)(4/5) = -9/5 + 4/5 = -5/5
Dᵤf(1,2) = -1
Problem 3: Maximum Rate of Change Intermediate
f(x,y) = e^(x² + y²)
Find the maximum rate of change at point (1,0) and the direction in which it occurs.
1 Compute gradient:
fₓ = 2x·e^(x²+y²), fᵧ = 2y·e^(x²+y²)
∇f = ⟨2x·e^(x²+y²), 2y·e^(x²+y²)⟩
2 Evaluate at (1,0):
∇f(1,0) = ⟨2(1)·e^(1), 2(0)·e^(1)⟩ = ⟨2e, 0⟩
3 Find maximum rate of change:
|∇f(1,0)| = √((2e)² + 0²) = 2e
4 Direction of maximum increase:
Direction = ∇f(1,0) = ⟨2e, 0⟩
Unit direction = ⟨1, 0⟩ (east direction)
Maximum rate: 2e, Direction: ⟨2e, 0⟩
Problem 4: Three Variables Gradient Advanced
f(x,y,z) = x²yz + sin(xz) + e^(xy)
Find ∇f at point (0,1,π/2).
1 Compute partial derivatives:
fₓ = 2xyz + z·cos(xz) + y·e^(xy)
fᵧ = x²z + x·e^(xy)
f_z = x²y + x·cos(xz)
2 Evaluate at (0,1,π/2):
fₓ(0,1,π/2) = 0 + (π/2)·cos(0) + 1·e⁰ = π/2 + 1
fᵧ(0,1,π/2) = 0 + 0·e⁰ = 0
f_z(0,1,π/2) = 0 + 0·cos(0) = 0
Correction: fᵧ = x²z + x·e^(xy) = 0 + 0·e⁰ = 0
Wait, check fᵧ at (0,1,π/2): 0²·(π/2) + 0·e⁰ = 0 + 0 = 0
Actually, fᵧ(0,1,π/2) = (0)²(π/2) + (0)e^(0) = 0
∇f(0,1,π/2) = ⟨π/2 + 1, 0, 0⟩
Problem 5: Tangent Plane using Gradient Advanced
f(x,y) = x² + xy + y²
Find the equation of the tangent plane to the surface at point (1,1,3).
1 Compute gradient:
∇f = ⟨2x + y, x + 2y⟩
∇f(1,1) = ⟨3, 3⟩
2 Equation of tangent plane:
z = f(1,1) + fₓ(1,1)(x-1) + fᵧ(1,1)(y-1)
f(1,1) = 1² + 1·1 + 1² = 3
3 Substitute values:
z = 3 + 3(x-1) + 3(y-1)
z = 3 + 3x - 3 + 3y - 3
z = 3x + 3y - 3
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🎓 Key Theoretical Concepts
Geometric Interpretation

• Gradient points in direction of steepest ascent

• |∇f| = maximum rate of change

• Gradient is perpendicular to level curves/surfaces

Directional Derivative Formula
Dᵤf(x₀,y₀) = lim_{h→0} [f(x₀+ha, y₀+hb) - f(x₀,y₀)]/h

where u = ⟨a,b⟩ is a unit vector

Applications

• Optimization problems

• Tangent planes to surfaces

• Gradient descent algorithms

• Physics: electric/magnetic fields