Question

Pls Help - Calc. HW dy/dx problem

Answer 1

`\displaystyle y' = (-2)/(x \ln (10)[\log (x) - 2]^2)`

General Formulas and Concepts:

Calculus

Differentiation

• Derivatives
• Derivative Notation

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

Derivative Rule [Basic Power Rule]:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

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

Explanation:

Step 1: Define

Identify.

`\displaystyle y = (\log (x))/(\log (x) - 2)`

Step 2: Differentiate

1. [Function] Derivative Rule [Quotient Rule]:
   `\displaystyle y' = ([\log (x) - 2][\log (x)]' - [\log (x) - 2]'[\log (x)])/([\log (x) - 2]^2)`
2. Rewrite [Derivative Rule - Addition/Subtraction]:
   `\displaystyle y' = ([\log (x) - 2][\log (x)]' - [\log (x)' - 2'][\log (x)])/([\log (x) - 2]^2)`
3. Logarithmic Differentiation:
   `\displaystyle y' = ([\log (x) - 2](1)/(\ln (10)x) - [(1)/(\ln (10)x) - 2'][\log (x)])/([\log (x) - 2]^2)`
4. Derivative Rule [Basic Power Rule]:
   `\displaystyle y' = ([\log (x) - 2](1)/(\ln (10)x) - (1)/(\ln (10)x)[\log (x)])/([\log (x) - 2]^2)`
5. Simplify:
   `\displaystyle y' = ((\log (x) - 2)/(\ln (10)x) - (\log (x))/(\ln (10)x))/([\log (x) - 2]^2)`
6. Simplify:
   `\displaystyle y' = ((-2)/(\ln (10)x))/([\log (x) - 2]^2)`
7. Rewrite:
   `\displaystyle y' = (-2)/(x \ln (10)[\log (x) - 2]^2)`

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

Unit: Differentiation