Answer:
(i) [tex]\displaystyle y' = (6x - 1)ln(2x + 1) + \frac{2x(3x - 1)}{2x + 1}[/tex]
(ii) [tex]\displaystyle y' = \frac{2x}{ln(x)} - \frac{x^2 + 2}{x(lnx)^2}[/tex]
(iii) [tex]\displaystyle y' = \frac{e^x[xln(2x) + 1]}{x}[/tex]
General Formulas and Concepts:
Calculus
Differentiation
- Derivatives
- Derivative Notation
Derivative Property [Multiplied Constant]: [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]
Derivative Property [Addition/Subtraction]: [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Derivative Rule [Product Rule]: [tex]\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)[/tex]
Derivative Rule [Quotient Rule]: [tex]\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}[/tex]
Derivative Rule [Chain Rule]: [tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]
Exponential Differentiation
Logarithmic Differentiation
Step-by-step explanation:
(i)
Step 1: Define
Identify
[tex]\displaystyle y = (3x^2 - x)ln(2x + 1)[/tex]
Step 2: Differentiate
- Product Rule: [tex]\displaystyle y' = (3x^2 - x)'ln(2x + 1) + (3x^2 - x)[ln(2x + 1)]'[/tex]
- Basic Power Rule/Logarithmic Differentiation [Chain Rule]: [tex]\displaystyle y' = (6x - 1)ln(2x + 1) + (3x^2 - x)\frac{1}{2x + 1}(2x + 1)'[/tex]
- Basic Power Rule: [tex]\displaystyle y' = (6x - 1)ln(2x + 1) + (3x^2 - x)\frac{2}{2x + 1}[/tex]
- Simplify [Factor]: [tex]\displaystyle y' = (6x - 1)ln(2x + 1) + \frac{2x(3x - 1)}{2x + 1}[/tex]
(ii)
Step 1: Define
Identify
[tex]\displaystyle y = \frac{x^2 + 2}{lnx}[/tex]
Step 2: Differentiate
- Quotient Rule: [tex]\displaystyle y' = \frac{(x^2 + 2)'lnx - (x^2 + 2)(lnx)'}{(lnx)^2}[/tex]
- Basic Power Rule/Logarithmic Differentiation: [tex]\displaystyle y' = \frac{2xlnx - (x^2 + 2)\frac{1}{x}}{(lnx)^2}[/tex]
- Rewrite: [tex]\displaystyle y' = \frac{2xlnx}{(lnx)^2} - \frac{(x^2 + 2)\frac{1}{x}}{(lnx)^2}[/tex]
- Simplify: [tex]\displaystyle y' = \frac{2x}{ln(x)} - \frac{x^2 + 2}{x(lnx)^2}[/tex]
(iii)
Step 1: Define
Identify
[tex]\displaystyle y = e^xln(2x)[/tex]
Step 2: Differentiate
- Product Rule: [tex]\displaystyle y' = (e^x)'ln(2x) + e^x[ln(2x)]'[/tex]
- Exponential Differentiation/Logarithmic Differentiation [Chain Rule]: [tex]\displaystyle y' = e^xln(2x) + e^x(\frac{1}{2x})(2x)'[/tex]
- Basic Power Rule: [tex]\displaystyle y' = e^xln(2x) + e^x(\frac{1}{2x})2[/tex]
- Simplify: [tex]\displaystyle y' = e^xln(2x) + \frac{e^x}{x}[/tex]
- Rewrite: [tex]\displaystyle y' = \frac{xe^xln(2x) + e^x}{x}[/tex]
- Factor: [tex]\displaystyle y' = \frac{e^x[xln(2x) + 1]}{x}[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Differentiation
Book: College Calculus 10e
