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Lapatulllka [165]
3 years ago
5

|5−4r|=17

Mathematics
2 answers:
kirill [66]3 years ago
8 0

Answer:

r=−3,11/2

Step-by-step explanation:

thats what I got

Brrunno [24]3 years ago
3 0

Answer:

R=-3 which is the same thing as 11/2

Step-by-step explanation:

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(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:

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative 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:

  1. f(x) = cxⁿ
  2. f’(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)

<u>Step 1: Define</u>

<em>Identify</em>

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

<u>Step 2: Differentiate</u>

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

(ii)

<u>Step 1: Define</u>

<em>Identify</em>

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

<u>Step 2: Differentiate</u>

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

(iii)

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle y = e^xln(2x)

<u>Step 2: Differentiate</u>

  1. Product Rule:                                                                                                 \displaystyle y' = (e^x)'ln(2x) + e^x[ln(2x)]'
  2. Exponential Differentiation/Logarithmic Differentiation [Chain Rule]:       \displaystyle y' = e^xln(2x) + e^x(\frac{1}{2x})(2x)'
  3. Basic Power Rule:                                                                                         \displaystyle y' = e^xln(2x) + e^x(\frac{1}{2x})2
  4. Simplify:                                                                                                         \displaystyle y' = e^xln(2x) + \frac{e^x}{x}
  5. Rewrite:                                                                                                         \displaystyle y' = \frac{xe^xln(2x) + e^x}{x}
  6. 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

6 0
3 years ago
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