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Help with num 1 please.​

Help with num 1 please.​-example-1
User Cmantas
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1 Answer

1 vote

Answer:

(i)
\displaystyle y' = (6x - 1)ln(2x + 1) + (2x(3x - 1))/(2x + 1)

(ii)
\displaystyle y' = (2x)/(ln(x)) - (x^2 + 2)/(x(lnx)^2)

(iii)
\displaystyle y' = (e^x[xln(2x) + 1])/(x)

General Formulas and Concepts:

Calculus

Differentiation

  • Derivatives
  • Derivative Notation

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

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

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]:
\displaystyle (d)/(dx) [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)

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

Derivative Rule [Chain Rule]:
\displaystyle (d)/(dx)[f(g(x))] =f'(g(x)) \cdot g'(x)

Exponential Differentiation

Logarithmic Differentiation

Explanation:

(i)

Step 1: Define

Identify


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

Step 2: Differentiate

  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)(1)/(2x + 1)(2x + 1)'
  3. Basic Power Rule:
    \displaystyle y' = (6x - 1)ln(2x + 1) + (3x^2 - x)(2)/(2x + 1)
  4. Simplify [Factor]:
    \displaystyle y' = (6x - 1)ln(2x + 1) + (2x(3x - 1))/(2x + 1)

(ii)

Step 1: Define

Identify


\displaystyle y = (x^2 + 2)/(lnx)

Step 2: Differentiate

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

(iii)

Step 1: Define

Identify


\displaystyle y = e^xln(2x)

Step 2: Differentiate

  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((1)/(2x))(2x)'
  3. Basic Power Rule:
    \displaystyle y' = e^xln(2x) + e^x((1)/(2x))2
  4. Simplify:
    \displaystyle y' = e^xln(2x) + (e^x)/(x)
  5. Rewrite:
    \displaystyle y' = (xe^xln(2x) + e^x)/(x)
  6. Factor:
    \displaystyle y' = (e^x[xln(2x) + 1])/(x)

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

Unit: Differentiation

Book: College Calculus 10e

User Johlo
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7.6k points