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

check above link.

graphically? well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.

Answer from: Quest

yes, gavin is right.

step-by-step explanation:

gavin is correct.

this problem is about converting 9% per year to how many percent per quarter. the rate is still the same. it only differed in the time period. so, through dimensional analysis,

9% per year × (1 year / 4 quarters) = 2.25% per quarter

if he was asked to convert 9% per year to how many percent per quarter compounded quarterly, then you will not simply divide by four. this is because compounding is not linear.

hope this , have a good day. c;

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Help with num 1 please.
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