Question

Help with num 1 please.​

Answer 1

(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