Techniques and Rules of Differentiation Study Guide

Differentiation of General Monomials

• Standard Rule: If $y = x^{n}$, the derivative is defined as:     $\frac{dy}{dx} = \frac{d}{dx}(x^{n}) = nx^{n-1}$

Differentiation of Polynomials

• Sum/Difference Rule: If $y = u + v + w + \dots$ where $u$, $v$, and $w$ are functions of $x$, then:     $\frac{dy}{dx} = \frac{d}{dx}(u + v + w + \dots) = \frac{du}{dx} + \frac{dv}{dx} + \frac{dw}{dx} + \dots$

• Constant Multiple Rule: If $y = cu$, where $c$ is a constant and $u$ is a function of $x$:     $\frac{d(cu)}{dx} = c \times \frac{du}{dx}$

Differentiation of Composite Functions

• Power of a Linear Function: If $y = (ax + b)^{n}$, the derivative is:     $\frac{dy}{dx} = \frac{d}{dx}(ax + b)^{n}$$\frac{dy}{dx} = n(ax + b)^{n-1} \times \frac{d}{dx}(ax + b)$

• Function of a Function (Chain Rule): If $y$ is a function of $u$, and $u$ is a function of $x$:     $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$

Product and Quotient Rules

• Product Rule (Two Functions): If $y = uv$, where $u$ and $v$ are functions of $x$:     $\frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx}$

• Product Rule (Three Functions): If $y = uvw$, where $u$, $v$, and $w$ are functions of $x$:     $\frac{dy}{dx} = vw\frac{du}{dx} + uw\frac{dv}{dx} + uv\frac{dw}{dx}$

• Quotient Rule: If $y = \frac{u}{v}$, where $u$ and $v$ are functions of $x$:     $\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^{2}}$

Exponential and Logarithmic Functions

• Exponential Function: If $y = e^{ax+b}$:     $\frac{dy}{dx} = \frac{d}{dx}(ax + b)e^{ax+b} = a e^{ax+b}$

• Logarithmic Function: If $y = \log(ax + b)$ or $y = \ln(ax + b)$, the derivative is:     $\frac{dy}{dx} = [\frac{1}{ax+b}] \times \frac{d}{dx}(ax + b) = \frac{a}{ax+b}$

Numerical Examples (Set 1)

• Example 20: Differentiate with respect to $x$:
  • (i) $y = 7x^{5} + 3x^{4} + x^{3} + 4$$\frac{dy}{dx} = 7\frac{d}{dx}(x^{5}) + 3\frac{d}{dx}(x^{4}) + \frac{d}{dx}(x^{3}) + \frac{d}{dx}(4)$$\frac{dy}{dx} = 7(5x^{4}) + 3(4x^{3}) + (3x^{2}) + 0 = 35x^{4} + 12x^{3} + 3x^{2}$
  • (ii) $y = \sqrt{x^{3}} - \sqrt[4]{x^{3}}$         Rewrite: $y = x^{\frac{3}{2}} - x^{\frac{3}{4}}$$\frac{dy}{dx} = \frac{3}{2}x^{(\frac{3}{2}-1)} - \frac{3}{4}x^{(\frac{3}{4}-1)}$$\frac{dy}{dx} = \frac{3}{2}x^{\frac{1}{2}} - \frac{3}{4}x^{-\frac{1}{4}}$
  • (iii) $y = \frac{4x^{3} + 3x^{2} + x}{x^{\frac{1}{3}}}$         Rewrite: $y = \frac{4x^{3}}{x^{\frac{1}{3}}} + \frac{3x^{2}}{x^{\frac{1}{3}}} + \frac{x}{x^{\frac{1}{3}}} = 4x^{\frac{8}{3}} + 3x^{\frac{5}{3}} + x^{\frac{2}{3}}$$\frac{dy}{dx} = 4(\frac{8}{3}x^{\frac{5}{3}}) + 3(\frac{5}{3}x^{\frac{2}{3}}) + \frac{2}{3}x^{-\frac{1}{3}}$$\frac{dy}{dx} = \frac{32}{3}x^{\frac{5}{3}} + 5x^{\frac{2}{3}} + \frac{2}{3\sqrt[3]{x}}$

Numerical Examples (Set 2)

• Example 21: Differentiate $y = (5x^{3} + 4)^{8}$ with respect to $x$:     Using Chain Rule:     $\frac{dy}{dx} = 8(5x^{3} + 4)^{7} \times \frac{d}{dx}(5x^{3} + 4)$$\frac{dy}{dx} = 8(5x^{3} + 4)^{7}(15x^{2}) = 120x^{2}(5x^{3} + 4)^{7}$

• Example 22: Differentiate $y = \frac{(x^{3}-1)^{4}}{(2x^{2}+1)^{4}}$ with respect to $x$:     Let $u = (x^{3}-1)^{4}$ and $v = (2x^{2}+1)^{4}$.     $\frac{du}{dx} = 4(x^{3}-1)^{3}(3x^{2}) = 12x^{2}(x^{3}-1)^{3}$$\frac{dv}{dx} = 4(2x^{2}+1)^{3}(4x) = 16x(2x^{2}+1)^{3}$     Applying Quotient Rule:     $\frac{dy}{dx} = \frac{(2x^{2}+1)^{4}[12x^{2}(x^{3}-1)^{3}] - (x^{3}-1)^{4}[16x(2x^{2}+1)^{3}]}{(2x^{2}+1)^{8}}$     Factor out $(2x^{2}+1)^{3}(x^{3}-1)^{3}$:     $\frac{dy}{dx} = \frac{(2x^{2}+1)^{3}(x^{3}-1)^{3}[12x^{2}(2x^{2}+1) - 16x(x^{3}-1)]}{(2x^{2}+1)^{8}}$$\frac{dy}{dx} = \frac{(x^{3}-1)^{3}[24x^{4} + 12x^{2} - 16x^{4} + 16x]}{(2x^{2}+1)^{5}}$$\frac{dy}{dx} = \frac{(x^{3}-1)^{3}[8x^{4} + 12x^{2} + 16x]}{(2x^{2}+1)^{5}} = \frac{4x(x^{3}-1)^{3}(2x^{3} + 3x + 4)}{(2x^{2}+1)^{5}}$

Numerical Examples (Set 3)

• Example 23: Product Rule Applications:

  • (i) $y = (x^{3} + 2)(x^{4} + 2x)$         Let $u = x^{3} + 2, \frac{du}{dx} = 3x^{2}$         Let $v = x^{4} + 2x, \frac{dv}{dx} = 4x^{3} + 2$$\frac{dy}{dx} = (x^{4} + 2x)(3x^{2}) + (x^{3} + 2)(4x^{3} + 2)$$\frac{dy}{dx} = 3x^{6} + 6x^{3} + 4x^{6} + 2x^{3} + 8x^{3} + 4$$\frac{dy}{dx} = 7x^{6} + 16x^{3} + 4$
  • (ii) $y = (x^{2} + x + 3)(2x^{2} + 3x + 1)$         Let $u = x^{2} + x + 3, \frac{du}{dx} = 2x + 1$         Let $v = 2x^{2} + 3x + 1, \frac{dv}{dx} = 4x + 3$$\frac{dy}{dx} = (2x^{2} + 3x + 1)(2x + 1) + (x^{2} + x + 3)(4x + 3)$$\frac{dy}{dx} = (4x^{3} + 2x^{2} + 6x^{2} + 3x + 2x + 1) + (4x^{3} + 3x^{2} + 4x^{2} + 3x + 12x + 9)$$\frac{dy}{dx} = 8x^{3} + 15x^{2} + 20x + 10$

• Example 24: Quotient Rule Application:     $y = \frac{4x^{2} + 3}{x^{2} + 1}$     Let $u = 4x^{2} + 3, \frac{du}{dx} = 8x$     Let $v = x^{2} + 1, \frac{dv}{dx} = 2x$$\frac{dy}{dx} = \frac{(x^{2} + 1)(8x) - (4x^{2} + 3)(2x)}{(x^{2} + 1)^{2}}$$\frac{dy}{dx} = \frac{8x^{3} + 8x - 8x^{3} - 6x}{(x^{2} + 1)^{2}} = \frac{2x}{(x^{2} + 1)^{2}}$

• Example 25: Quotient Rule Application:     $y = \frac{2x - 3}{3x + 4}$     Let $u = 2x - 3, \frac{du}{dx} = 2$     Let $v = 3x + 4, \frac{dv}{dx} = 3$$\frac{dy}{dx} = \frac{(3x + 4)(2) - (2x - 3)(3)}{(3x + 4)^{2}} = \frac{6x + 8 - 6x + 9}{(3x + 4)^{2}} = \frac{17}{(3x + 4)^{2}}$

General Examples (Page 8-11)

• (i) Derivative of $y = \frac{\sqrt{x-1}}{\sqrt{x+1}}$:     Rewrite: $y = \frac{(x-1)^{\frac{1}{2}}}{(x+1)^{\frac{1}{2}}}$     Result calculation (via Quotient rule simplification):     $\frac{dy}{dx} = \frac{1}{(x+1)^{\frac{3}{2}}(x-1)^{\frac{1}{2}}}$

• (ii) Derivative of $y = (x-1)\sqrt{x^{2} - 2x + 2}$:     Let $u = x - 1, \frac{du}{dx} = 1$     Let $v = (x^{2} - 2x + 2)^{\frac{1}{2}}, \frac{dv}{dx} = \frac{1}{2}(x^{2} - 2x + 2)^{-\frac{1}{2}}(2x - 2) = (x-1)(x^{2}-2x+2)^{-\frac{1}{2}}$$\frac{dy}{dx} = (x^{2}-2x+2)^{\frac{1}{2}} + (x-1)^{2}(x^{2}-2x+2)^{-\frac{1}{2}}$$\frac{dy}{dx} = \frac{x^{2} - 2x + 2 + (x^{2} - 2x + 1)}{\sqrt{x^{2} - 2x + 2}} = \frac{2x^{2} - 4x + 3}{\sqrt{x^{2} - 2x + 2}}$

• (iii) Derivative of $y = (x^{2}-1)^{3}(5x^{3}+2)^{4}$:     Using Product Rule:     $\frac{dy}{dx} = 6x(5x^{3}+2)^{3}(x^{2}-1)^{2}[15x^{3}-10x+2]$

• (iv) Derivative of $y = \sqrt{1 + \sqrt{x}}$:     Rewrite: $y = (1 + x^{\frac{1}{2}})^{\frac{1}{2}}$$\frac{dy}{dx} = \frac{1}{2}(1 + x^{\frac{1}{2}})^{-\frac{1}{2}} \times (\frac{1}{2}x^{-\frac{1}{2}}) = \frac{1}{4\sqrt{x}\sqrt{1+\sqrt{x}}}$

Exponential and Logarithmic Detailed Examples

• Example 26: $y = x^{3}e^{x^{2}}$$u = x^{3}, u' = 3x^{2}$$v = e^{x^{2}}, v' = 2xe^{x^{2}}$$\frac{dy}{dx} = e^{x^{2}}(3x^{2}) + x^{3}(2xe^{x^{2}}) = x^{2}e^{x^{2}}(3 + 2x^{2})$

• Example 27: $y = (4x^{2} + 3)^{3}e^{(3x^{2} + 2)^{2}}$     After simplification using product and chain rules:     $\frac{dy}{dx} = 12x(4x^{2} + 3)^{2}e^{(3x^{2}+2)^{2}}[12x^{4} + 17x^{2} + 8]$

• Example 28: $y = x \ln(x)$$\frac{dy}{dx} = \ln(x)(1) + x(\frac{1}{x}) = \ln(x) + 1$

• Example 29: $y = e^{2x}\ln(x^{2} + 1)$$\frac{dy}{dx} = 2e^{2x}\ln(x^{2}+1) + e^{2x}(\frac{2x}{x^{2}+1}) = 2e^{2x}[\ln(x^{2}+1) + \frac{x}{x^{2}+1}]$

• Example 30: Derivative of $y = x^{x}$:     Take natural log: $\ln(y) = x\ln(x)$     Differentiate implicitly: $\frac{1}{y}\frac{dy}{dx} = \ln(x) + 1$$\frac{dy}{dx} = y(\ln(x) + 1) = x^{x}(\ln(x) + 1)$

• Example 31: $y = 8^{x}$$\ln(y) = x\ln(8)$$\frac{1}{y}\frac{dy}{dx} = \ln(8)$$\frac{dy}{dx} = 8^{x}\ln(8)$

• Example 32: $y = a^{\cos(x)}$$\frac{dy}{dx} = -a^{\cos(x)}\ln(a)\sin(x)$

Implicit Function Differentiation

• Definition: If $f(x, y) = 0$, then $y$ is defined implicitly as a function of $x$.

• Example 33: Find $\frac{dy}{dx}$ for $xy + x - 2y - 1 = 0$$\frac{d}{dx}(xy) + \frac{d}{dx}(x) - \frac{d}{dx}(2y) - \frac{d}{dx}(1) = 0$$(y + x\frac{dy}{dx}) + 1 - 2\frac{dy}{dx} = 0$$\frac{dy}{dx}(x - 2) = -y - 1$$\frac{dy}{dx} = \frac{y + 1}{2 - x}$

• Example 34: Find $\frac{dy}{dx}$ for $e^{x+y} = 5xy$$e^{x+y}(1 + \frac{dy}{dx}) = 5(y + x\frac{dy}{dx})$$e^{x+y} + e^{x+y}\frac{dy}{dx} = 5y + 5x\frac{dy}{dx}$$\frac{dy}{dx}(e^{x+y} - 5x) = 5y - e^{x+y}$$\frac{dy}{dx} = \frac{5y - e^{x+y}}{e^{x+y} - 5x}$

• Example 35: Find $\frac{dy}{dx}$ at point $(2, 3)$ for $3x^{2} + 2y^{2} + xy^{2} + x - 7 = 0$$6x + 4y\frac{dy}{dx} + (y^{2} + 2xy\frac{dy}{dx}) + 1 = 0$$\frac{dy}{dx}(4y + 2xy) = -(1 + 6x + y^{2})$     At $(2, 3)$: $\frac{dy}{dx}(4(3) + 2(2)(3)) = -(1 + 6(2) + 3^{2})$$\frac{dy}{dx}(12 + 12) = -(1 + 12 + 9) \implies 24\frac{dy}{dx} = -22 \implies \frac{dy}{dx} = -\frac{11}{12}$

Trigonometric Functions

• Basic Derivatives:

  • (i) $y = \sin(x) \implies \frac{dy}{dx} = \cos(x)$
  • (ii) $y = \cos(x) \implies \frac{dy}{dx} = -\sin(x)$
  • (iii) $y = \tan(x) \implies \frac{dy}{dx} = \sec^{2}(x)$
  • (iv) $y = \sec(x) \implies \frac{dy}{dx} = \sec(x)\tan(x)$
  • (v) $y = \cot(x) \implies \frac{dy}{dx} = -\csc^{2}(x)$
  • (vi) $y = \csc(x) \implies \frac{dy}{dx} = -\csc(x)\cot(x)$

• Example 40: Differentiate $y = \cot(3x)$     Using chain rule: $\frac{dy}{dx} = -3\csc^{2}(3x)$

• Example 41: Differentiate $y = \frac{x}{\tan(x)}$ or $y = x\cot(x)$     Using Product Rule on $x\cot(x)$:     $\frac{dy}{dx} = \cot(x) - x\csc^{2}(x)$

Parametric Functions

• Rule: If $x = f(t)$ and $y = g(t)$, then:     $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

• Example 43: Find $\frac{dy}{dx}$ when $y = t^{2}$ and $x = \frac{1}{t}$$\frac{dy}{dt} = 2t$$\frac{dx}{dt} = -t^{-2}$$\frac{dy}{dx} = \frac{2t}{-t^{-2}} = -2t^{3}$

• Example 44: If $x = t^{3} + t, y = 2t^{2}$, find $\frac{dy}{dx}$ at $t = 1$$\frac{dy}{dt} = 4t, \frac{dx}{dt} = 3t^{2} + 1$$\frac{dy}{dx} = \frac{4t}{3t^{2} + 1}$     At $t = 1$: $\frac{4(1)}{3(1)^{2} + 1} = \frac{4}{4} = 1$

Inverse Trigonometric Functions

• Formula Derivations:

  • If $y = \sin^{-1}(x)$, then $\frac{dy}{dx} = \frac{1}{\sqrt{1-x^{2}}}$
  • If $y = \cos^{-1}(x)$, then $\frac{dy}{dx} = -\frac{1}{\sqrt{1-x^{2}}}$
  • If $y = \tan^{-1}(x)$, then $\frac{dy}{dx} = \frac{1}{1+x^{2}}$

• Example 46: $y = \sin^{-1}(2x-1)$$\frac{dy}{dx} = \frac{2}{\sqrt{1 - (2x-1)^{2}}} = \frac{2}{\sqrt{4x - 4x^{2}}} = \frac{1}{\sqrt{x-x^{2}}}$

• Example 47: $y = \tan^{-1}(\frac{1+x}{1-x})$     Using chain rule and quotient rule on the argument:     $\frac{dy}{dx} = \frac{1}{1+x^{2}}$

• Example 48: $y = x^{2}\cos^{-1}(2x^{-1})$     Using Product Rule:     $\frac{dy}{dx} = 2x\cos^{-1}(\frac{2}{x}) + \frac{2}{\sqrt{1 - \frac{4}{x^{2}}}}$$\frac{dy}{dx} = 2x\cos^{-1}(\frac{2}{x}) + \frac{2x}{\sqrt{x^{2}-4}}$

Exercises Summary

• Exercise 1.1: Evaluate various limits involving polynomial fractions, roots, exponentials, and logarithms (28 items).
• Exercise 1.2: Differentiate from 1st principles (e.g., $y = x^{2}+2, y = \sin(2x), y = \sqrt{2x+1}$ etc.).
• Exercise 1.3: Differentiate complex expressions involving products, quotients, inverse trig, and implicit functions (40 items).