Home Class 11 Applied Maths Chapter 7: Calculus Exercise 7.4: Rules and Formulas of Differentiation
Chapter 7 • Unit III

Exercise 7.4: Rules and Formulas of Differentiation

Comprehensive guide and practice questions for CBSE Class 11 Applied Mathematics curriculum (2026-27).

7.4 Rules and Formulas of Differentiation

In Applied Mathematics, corporate financial engineering, and revenue management, functions rarely exist in isolation; they appear as products, quotients, and nested compositions. The rules of differentiation—Power Rule, Product Rule, Quotient Rule, Chain Rule, and Logarithmic Differentiation—provide the algorithmic engine to differentiate complex economic models rapidly and accurately.

Master Formulary of Differentiation Rules

  • Power Rule: \[ \frac{d}{dx}(x^n) = n x^{n-1} \quad (\text{for any real } n) \]
  • Constant Multiple Rule: \[ \frac{d}{dx}[c \cdot f(x)] = c \cdot f'(x) \]
  • Sum & Difference Rule: \[ \frac{d}{dx}[u(x) \pm v(x)] = u'(x) \pm v'(x) \]
  • Product Rule (Leibniz Rule): \[ \frac{d}{dx}[u \cdot v] = u \frac{dv}{dx} + v \frac{du}{dx} = u v' + v u' \] For three factors: \((u v w)' = u' v w + u v' w + u v w'\).
  • Quotient Rule: \[ \frac{d}{dx} \left[ \frac{u}{v} \right] = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} = \frac{v u' - u v'}{v^2} \quad (v \neq 0) \]
  • Chain Rule (Composite Functions): If \(y = f(u)\) and \(u = g(x)\): \[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = f'(g(x)) \cdot g'(x) \] Generalized Power Rule: \(\frac{d}{dx}[u(x)]^n = n [u(x)]^{n-1} \cdot u'(x)\).
  • Exponential & Logarithmic Derivatives (Applied Math): \[ \frac{d}{dx}(e^x) = e^x, \quad \frac{d}{dx}(e^{kx}) = k e^{kx}, \quad \frac{d}{dx}(a^x) = a^x \ln a \] \[ \frac{d}{dx}(\ln x) = \frac{1}{x}, \quad \frac{d}{dx}(\log_a x) = \frac{1}{x \ln a} \]
  • Second Order Derivative (Concavity & Acceleration): \[ \frac{d^2 y}{dx^2} = f''(x) = \frac{d}{dx} \left( \frac{dy}{dx} \right) \]

Systematic Problem-Solving Strategy

Form of Expression Primary Rule to Deploy Algebraic Tip
\(y = (2x + 3)^5\) Chain Rule (Generalized Power) Differentiate outer power, multiply by inner derivative: \(5(2x+3)^4 \times 2 = 10(2x+3)^4\)
\(y = x^2 e^{3x}\) Product Rule + Exponential Chain \(x^2 \frac{d}{dx}(e^{3x}) + e^{3x} \frac{d}{dx}(x^2) = x^2(3e^{3x}) + 2x e^{3x}\)
\(y = \frac{2x - 1}{3x + 4}\) Quotient Rule Denominator squared on bottom; keep \(v u' - u v'\) in correct order
\(y = x^x\) Logarithmic Differentiation Take \(\ln\) on both sides: \(\ln y = x \ln x\), then differentiate implicitly

Comprehensive Multi-Step Worked Examples

Differentiate the following with respect to \(x\):

  1. \(\displaystyle y = \frac{x^2 + 1}{x^2 - 1}\)
  2. \(\displaystyle y = (3x^2 - 5)^4 \cdot (2x + 1)^3\)
  3. \(\displaystyle y = e^{2x} \ln(x^2 + 1)\)

Solution:

  1. Quotient Rule: \[ \frac{dy}{dx} = \frac{(x^2 - 1) \frac{d}{dx}(x^2 + 1) - (x^2 + 1) \frac{d}{dx}(x^2 - 1)}{(x^2 - 1)^2} \] \[ = \frac{(x^2 - 1)(2x) - (x^2 + 1)(2x)}{(x^2 - 1)^2} = \frac{2x^3 - 2x - (2x^3 + 2x)}{(x^2 - 1)^2} = \frac{-4x}{(x^2 - 1)^2}. \]
  2. Product & Chain Rules: \[ \frac{dy}{dx} = (3x^2 - 5)^4 \frac{d}{dx}[(2x + 1)^3] + (2x + 1)^3 \frac{d}{dx}[(3x^2 - 5)^4] \] \[ = (3x^2 - 5)^4 [3(2x + 1)^2 \times 2] + (2x + 1)^3 [4(3x^2 - 5)^3 \times 6x] \] \[ = 6(3x^2 - 5)^4 (2x + 1)^2 + 24x(2x + 1)^3 (3x^2 - 5)^3 \] Factor out common terms \(6(3x^2 - 5)^3 (2x + 1)^2\): \[ = 6(3x^2 - 5)^3 (2x + 1)^2 [(3x^2 - 5) + 4x(2x + 1)] \] \[ = 6(3x^2 - 5)^3 (2x + 1)^2 [3x^2 - 5 + 8x^2 + 4x] = 6(3x^2 - 5)^3 (2x + 1)^2 (11x^2 + 4x - 5). \]
  3. Product & Chain on Exponential & Logarithmic: \[ \frac{dy}{dx} = e^{2x} \frac{d}{dx}[\ln(x^2 + 1)] + \ln(x^2 + 1) \frac{d}{dx}[e^{2x}] \] \[ = e^{2x} \left[ \frac{2x}{x^2 + 1} \right] + 2 e^{2x} \ln(x^2 + 1) = 2 e^{2x} \left[ \frac{x}{x^2 + 1} + \ln(x^2 + 1) \right]. \]
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