Stewart Calculus · 8th Edition
Chapter 14: Partial Derivatives
Every key formula from Stewart Calculus Chapter 14, in one searchable page. Click a card to study it — worked examples included.
14.1–14.3 · FUNCTIONS & PARTIALS
Level curves: f(x, y) = k.
fx: hold y still · fy: hold x still
Clairaut: fxy = fyx (continuous partials).
Formal definition
fx(a,b) = limh→0 f(a+h,b) − f(a,b)h
Slope of the trace: slice the surface with y = b.
Higher-order partials
fxyy = fyxy = fyyx (any order, continuous)
Synonyms: fx = ∂f/∂x = Dxf.
WORKED EXAMPLE
f = x³y⁴: fx = 3x²y⁴, fy = 4x³y³;
at (1, −1): fx = 3, fy = −4.
at (1, −1): fx = 3, fy = −4.
14.4 · TANGENT PLANES & APPROX
z − z₀ = fx(x₀,y₀)(x−x₀) + fy(x₀,y₀)(y−y₀)
Linear approximation & differentials
L(x,y) = f(x₀,y₀) + fxΔx + fyΔy
dz = fx dx + fy dy
Differentiability
fx, fy exist & continuous near (a,b) ⇒ differentiable:
Δz = fxΔx + fyΔy + ε₁Δx + ε₂Δy, ε₁, ε₂ → 0.
Δz = fxΔx + fyΔy + ε₁Δx + ε₂Δy, ε₁, ε₂ → 0.
Increment vs differential
The error |Δz − dz| shrinks faster than Δx, Δy.
WORKED EXAMPLE
Tangent plane to z = x² − xy + y² at (1, −1, 3):
fx = 2x − y = 3, fy = −x + 2y = −3 ⇒
z − 3 = 3(x − 1) − 3(y + 1).
fx = 2x − y = 3, fy = −x + 2y = −3 ⇒
z − 3 = 3(x − 1) − 3(y + 1).
14.5 · CHAIN RULE
dz/dt = fx x′ + fy y′
∂z/∂s = fx xs + fy ys
Implicit F(x,y) = 0: dy/dx = −Fx/Fy.
WORKED EXAMPLE
z = x²y, x = t, y = t²:
dz/dt = 2xy · 1 + x² · 2t = 4t³ ⇒ z = t⁴ ✓
dz/dt = 2xy · 1 + x² · 2t = 4t³ ⇒ z = t⁴ ✓
Chain rule (tree)
Branches: x(t), y(t) ⇒ dz/dt = zxx′ + zyy′.
Implicit (three variables)
F(x,y,z) = 0 ⇒ ∂z∂x = −Fx/Fz, ∂z∂y = −Fy/Fz (Fz ≠ 0)
WORKED EXAMPLE
z = eu sin v, u = st², v = s² + t:
∂z/∂s = t²eu sin v + 2seu cos v.
∂z/∂s = t²eu sin v + 2seu cos v.
14.6 · GRADIENT
∇f = ⟨fx, fy⟩, Duf = ∇f · u
Steepest ascent: |∇f| in direction ∇f.
∇f ⊥ level curves.
∇f ⊥ level curves.
WORKED EXAMPLE
f = x²y at (1, 2), u = ⟨3/5, 4/5⟩:
∇f = ⟨4, 1⟩ ⇒ Duf = 16/5.
∇f = ⟨4, 1⟩ ⇒ Duf = 16/5.
Max / min rate
max Duf = |∇f| (dir ∇f) · min = −|∇f| (dir −∇f)
Duf = 0 when u ⊥ ∇f (tangent to the level curve).
Tangent plane to a level surface
F(x,y,z) = k: Fx(x₀)(x−x₀) + Fy(y₀)(y−y₀) + Fz(z₀)(z−z₀) = 0
Normal line: x−x₀Fx = y−y₀Fy = z−z₀Fz.
WORKED EXAMPLE
f = x²y at (1, 2): fastest increase in dir ⟨4, 1⟩;
max rate = |∇f| = √17.
max rate = |∇f| = √17.
14.7–14.8 · MAX/MIN & LAGRANGE
Second Derivatives Test
D = fxxfyy − (fxy)²
D > 0, fxx > 0 min · D > 0, fxx < 0 max · D < 0 saddle
Lagrange multipliers
∇f = λ∇g subject to g = k
WORKED EXAMPLES
f = x³ + y² − 2xy: crit (0,0), (2/3, 2/3);
D = 12x − 4: saddle at (0,0), local min at (2/3, 2/3).
Max xy on x²+y² = 1: 1/2 at (1/√2, 1/√2).
D = 12x − 4: saddle at (0,0), local min at (2/3, 2/3).
Max xy on x²+y² = 1: 1/2 at (1/√2, 1/√2).
Absolute max / min (closed, bounded D)
Candidates: interior critical points + boundary values — compare all.
Two constraints
∇f = λ∇g + μ∇h subject to g = k, h = l
WORKED EXAMPLE
Abs max/min of xy on x² + y² ≤ 4:
interior (0,0) → 0; boundary 2 sin 2t ⇒
max 2 at (√2, √2), min −2.
interior (0,0) → 0; boundary 2 sin 2t ⇒
max 2 at (√2, √2), min −2.
WATCH OUT!
Mistakes that cost points
- Clairaut needs CONTINUOUS mixed partials
- D = 0 ⇒ the test says nothing!
- Duf needs a UNIT vector u
- Lagrange also requires ∇g ≠ 0
- Saddle ≠ min/max — check the sign of D
- Tangent plane needs fx, fy AT the point
- |∇f| is the rate; ∇f/|∇f| is the direction
- Second partials: differentiate in the RIGHT order
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