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Commission Brokerage and Discount

Maharashtra Board · Class 12 · Mathematics & Statistics -Commerce

Flashcards for Commission Brokerage and Discount — Maharashtra Board Class 12 Mathematics & Statistics -Commerce. Quick Q&A cards covering key concepts, definitions, and formulas.

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Card 1Composite Functions

What is the Chain Rule for differentiating composite functions? State the formula and explain when to use it.

Answer

Chain Rule Formula: If y = f(u) and u = g(x), then dy/dx = (dy/du) × (du/dx) Alternatively: If y = f[g(x)], then dy/dx = f'[g(x)] × g'(x) Use when: Differentiating functions within functions, like s

Card 2Composite Functions

Differentiate y = √(x² + 5) using the chain rule. Show all steps.

Answer

Step 1: Identify functions - Outer function: √u where u = x² + 5 - Inner function: u = x² + 5 Step 2: Find derivatives - dy/du = 1/(2√u) = 1/(2√(x² + 5)) - du/dx = 2x Step 3: Apply chain rule dy/dx

Card 3Inverse Functions

What is the formula for differentiating inverse functions? State the theorem and explain the geometric meaning.

Answer

Inverse Function Theorem: If y = f(x) is differentiable and dy/dx ≠ 0, and x = f⁻¹(y) exists, then: dx/dy = 1/(dy/dx) Geometric Meaning: - The slope of f⁻¹ at point (a,b) is the reciprocal of the sl

Card 4Inverse Functions

Find the derivative of sin⁻¹(x). Derive it step by step using the inverse function theorem.

Answer

Given: y = sin⁻¹(x), where -1 ≤ x ≤ 1, -π/2 ≤ y ≤ π/2 Step 1: Express inverse relationship If y = sin⁻¹(x), then x = sin y Step 2: Differentiate x = sin y with respect to y dx/dy = cos y Step 3: Ap

Card 5Inverse Functions

List the derivatives of all six inverse trigonometric functions with their domains.

Answer

1. d/dx[sin⁻¹(x)] = 1/√(1-x²), |x| < 1 2. d/dx[cos⁻¹(x)] = -1/√(1-x²), |x| < 1 3. d/dx[tan⁻¹(x)] = 1/(1+x²), x ∈ ℝ 4. d/dx[cot⁻¹(x)] = -1/(1+x²), x ∈ ℝ 5. d/dx[sec⁻¹(x)] = 1/(x√(x²-1)), |x| > 1

Card 6Inverse Functions

Differentiate y = tan⁻¹(2x/(1-x²)) and simplify your answer.

Answer

Method 1: Direct differentiation dy/dx = 1/(1+(2x/(1-x²))²) × d/dx[2x/(1-x²)] Using quotient rule for 2x/(1-x²): d/dx[2x/(1-x²)] = [2(1-x²) - 2x(-2x)]/(1-x²)² = (2+2x²)/(1-x²)² Therefore: dy/dx = 2/

Card 7Logarithmic Differentiation

When should you use logarithmic differentiation? List the conditions and explain the process.

Answer

Use logarithmic differentiation when: 1. Function has form y = [f(x)]^g(x) (variable base and exponent) 2. Products/quotients with multiple terms and powers 3. Complex expressions with roots, powers,

Card 8Logarithmic Differentiation

Differentiate y = x^x using logarithmic differentiation. Show complete steps.

Answer

Step 1: Take natural log of both sides ln y = ln(x^x) = x ln x Step 2: Differentiate both sides with respect to x d/dx(ln y) = d/dx(x ln x) Step 3: Left side using chain rule (1/y)(dy/dx) = d/dx(x l

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What are the important topics in Commission Brokerage and Discount for Maharashtra Board Class 12 Mathematics & Statistics -Commerce?
Commission Brokerage and Discount covers several key topics that are frequently asked in Maharashtra Board Class 12 board exams. Focus on the core concepts listed on this page and practise related questions to build confidence.
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