Differentiation

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If y = f(u) and u = g(x), then: dy/dx = (dy/du) × (du/dx) This is used when we have a function inside another function. Example: For y = (3x² + 5)⁴ Let u = 3x² + 5, so y = u⁴ dy/du = 4u³, du/dx = 6x

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Step-by-step solution: Step 1: Identify the composite function Let u = 4x³ + 3x² - 2x, so y = u⁶ Step 2: Find derivatives dy/du = 6u⁵ du/dx = 12x² + 6x - 2 Step 3: Apply chain rule dy/dx = (dy/du) ×

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If y = f(x) has an inverse function x = f⁻¹(y), then: dx/dy = 1/(dy/dx), provided dy/dx ≠ 0 This means the derivative of the inverse function is the reciprocal of the derivative of the original funct

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Step-by-step solution: Step 1: Find dy/dx dy/dx = 15 + 2x Step 2: Apply inverse function rule dx/dy = 1/(dy/dx) dx/dy = 1/(15 + 2x) Step 3: Interpret Rate of change of demand with respect to price =

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Logarithmic differentiation is used for functions involving: 1. Products of multiple functions 2. Quotients of complex functions 3. Functions with variable exponents [f(x)]^g(x) Method: 1. Take natur

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Step-by-step solution: Step 1: Take natural logarithm ln y = ln[(3 + x)^x] = x ln(3 + x) Step 2: Differentiate both sides (1/y) × (dy/dx) = x × 1/(3 + x) + ln(3 + x) × 1 (1/y) × (dy/dx) = x/(3 + x) +

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Step-by-step solution: Step 1: Rewrite using negative exponents y = (2x + 3)⁵(3x - 1)⁻³(5x - 2)⁻¹/² Step 2: Take natural logarithm ln y = 5ln(2x + 3) - 3ln(3x - 1) - (1/2)ln(5x - 2) Step 3: Differen

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Implicit Function: A relation where y cannot be easily expressed as y = f(x) Examples: x² + y² = 25, xy = log(xy) Differentiation Process: 1. Differentiate both sides with respect to x 2. Use chain r