Answer:
[tex]\displaystyle y' = \frac{-2}{x \ln (10)[\log (x) - 2]^2}[/tex]
General Formulas and Concepts:
Calculus
Differentiation
- Derivatives
- Derivative Notation
Derivative Property [Addition/Subtraction]: [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]
Derivative Rule [Basic Power Rule]:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Derivative Rule [Quotient Rule]: [tex]\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}[/tex]
Step-by-step explanation:
Step 1: Define
Identify.
[tex]\displaystyle y = \frac{\log (x)}{\log (x) - 2}[/tex]
Step 2: Differentiate
- [Function] Derivative Rule [Quotient Rule]: [tex]\displaystyle y' = \frac{[\log (x) - 2][\log (x)]' - [\log (x) - 2]'[\log (x)]}{[\log (x) - 2]^2}[/tex]
- Rewrite [Derivative Rule - Addition/Subtraction]: [tex]\displaystyle y' = \frac{[\log (x) - 2][\log (x)]' - [\log (x)' - 2'][\log (x)]}{[\log (x) - 2]^2}[/tex]
- Logarithmic Differentiation: [tex]\displaystyle y' = \frac{[\log (x) - 2]\frac{1}{\ln (10)x} - [\frac{1}{\ln (10)x} - 2'][\log (x)]}{[\log (x) - 2]^2}[/tex]
- Derivative Rule [Basic Power Rule]: [tex]\displaystyle y' = \frac{[\log (x) - 2]\frac{1}{\ln (10)x} - \frac{1}{\ln (10)x}[\log (x)]}{[\log (x) - 2]^2}[/tex]
- Simplify: [tex]\displaystyle y' = \frac{\frac{\log (x) - 2}{\ln (10)x} - \frac{\log (x)}{\ln (10)x}}{[\log (x) - 2]^2}[/tex]
- Simplify: [tex]\displaystyle y' = \frac{\frac{-2}{\ln (10)x}}{[\log (x) - 2]^2}[/tex]
- Rewrite: [tex]\displaystyle y' = \frac{-2}{x \ln (10)[\log (x) - 2]^2}[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Differentiation
