Differentials and Partial Derivatives

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Step 1: Find f(x₀) = f(3) = √(1 + 3) = √4 = 2 Step 2: Find f'(x) = 1/(2√(1 + x)), so f'(3) = 1/(2√4) = 1/4 Step 3: Linear approximation L(x) = f(x₀) + f'(x₀)(x - x₀) = 2 + (1/4)(x - 3) Step 4: Estimat…

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Step 1: Choose f(x) = √x, x₀ = 9 (nearest perfect square), Δx = 0.2 Step 2: Find f(9) = √9 = 3 Step 3: Find f'(x) = 1/(2√x), so f'(9) = 1/(2√9) = 1/6 Step 4: Apply formula: f(x₀ + Δx) ≈ f(x₀) + f'(x₀)…

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Type of Error: When using linear approximation L(x) instead of actual value f(x), the error is E = |f(x) - L(x)| Error Classifications: 1. Absolute Error = |Actual Value - Approximate Value| 2. Relati…

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Step 1: Formula for area = A = πr² Step 2: Actual area = π(12.5)² = 156.25π cm² Step 3: Approximate area = π(12.65)² = 160.0225π cm² Step 4: Absolute Error = |156.25π - 160.0225π| = 3.7725π ≈ 11.85 cm…

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Formula: df = f'(x)dx or dy = f'(x)dx where dx = Δx (change in x), f'(x) is the derivative What it represents: The approximate change in f along the tangent line when x changes by dx When to use it: 1…

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Step 1: Find f'(x) = d/dx(x³ - 2x² + 5x) = 3x² - 4x + 5 Step 2: Evaluate f'(2) = 3(2)² - 4(2) + 5 = 12 - 8 + 5 = 9 Step 3: Apply differential formula: df = f'(x)dx = 9(0.1) = 0.9 Step 4: Verify by cal…

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Step 1: Volume formula for sphere: V = (4/3)πr³ Step 2: Find dV/dr = 4πr² Step 3: At r = 10: dV/dr = 4π(10)² = 400π Step 4: Change in radius: dr = -0.1 cm (negative because it decreases) Step 5: Apply…

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Partial Derivative with respect to x: ∂F/∂x(x₀, y₀) = lim(h→0) [F(x₀+h, y₀) - F(x₀, y₀)]/h Partial Derivative with respect to y: ∂F/∂y(x₀, y₀) = lim(k→0) [F(x₀, y₀+k) - F(x₀, y₀)]/k Key Concept: When…