Mathematics, 16.07.2021 14:00 flex23

Help with num 1 please.

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Answer from: monique69

(i) $\displaystyle y' = (6x - 1)ln(2x + 1) + \frac{2x(3x - 1)}{2x + 1}$

(ii) $\displaystyle y' = \frac{2x}{ln(x)} - \frac{x^2 + 2}{x(lnx)^2}$

(iii) $\displaystyle y' = \frac{e^x[xln(2x) + 1]}{x}$

General Formulas and Concepts:

Calculus

Differentiation

DerivativesDerivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Derivative Rule [Quotient Rule]: $\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Exponential Differentiation

Logarithmic Differentiation

Step-by-step explanation:

(i)

Step 1: Define

Identify

$\displaystyle y = (3x^2 - x)ln(2x + 1)$

Step 2: Differentiate

Product Rule: $\displaystyle y' = (3x^2 - x)'ln(2x + 1) + (3x^2 - x)[ln(2x + 1)]'$ Basic Power Rule/Logarithmic Differentiation [Chain Rule]: $\displaystyle y' = (6x - 1)ln(2x + 1) + (3x^2 - x)\frac{1}{2x + 1}(2x + 1)'$ Basic Power Rule: $\displaystyle y' = (6x - 1)ln(2x + 1) + (3x^2 - x)\frac{2}{2x + 1}$ Simplify [Factor]: $\displaystyle y' = (6x - 1)ln(2x + 1) + \frac{2x(3x - 1)}{2x + 1}$

(ii)

Step 1: Define

Identify

$\displaystyle y = \frac{x^2 + 2}{lnx}$

Step 2: Differentiate

Quotient Rule: $\displaystyle y' = \frac{(x^2 + 2)'lnx - (x^2 + 2)(lnx)'}{(lnx)^2}$ Basic Power Rule/Logarithmic Differentiation: $\displaystyle y' = \frac{2xlnx - (x^2 + 2)\frac{1}{x}}{(lnx)^2}$ Rewrite: $\displaystyle y' = \frac{2xlnx}{(lnx)^2} - \frac{(x^2 + 2)\frac{1}{x}}{(lnx)^2}$ Simplify: $\displaystyle y' = \frac{2x}{ln(x)} - \frac{x^2 + 2}{x(lnx)^2}$

(iii)

Step 1: Define

Identify

$\displaystyle y = e^xln(2x)$

Step 2: Differentiate

Product Rule: $\displaystyle y' = (e^x)'ln(2x) + e^x[ln(2x)]'$ Exponential Differentiation/Logarithmic Differentiation [Chain Rule]: $\displaystyle y' = e^xln(2x) + e^x(\frac{1}{2x})(2x)'$ Basic Power Rule: $\displaystyle y' = e^xln(2x) + e^x(\frac{1}{2x})2$ Simplify: $\displaystyle y' = e^xln(2x) + \frac{e^x}{x}$ Rewrite: $\displaystyle y' = \frac{xe^xln(2x) + e^x}{x}$ Factor: $\displaystyle y' = \frac{e^x[xln(2x) + 1]}{x}$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

Book: College Calculus 10e

Answer from: Quest

the answer to this question is yes

Answer from: Quest

step-by-step explanation:

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