Answer:
\(\displaystyle \frac{dy}{dx} = 1 - \frac{\ln x}{\big( \ln (x) + 1 \big)^2}\)
General Formulas and Concepts:
Calculus
Differentiation
- Derivatives
- Derivative Notation
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:
- f(x) = cxⁿ
- 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]: \(\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}\)
Step-by-step explanation:
Step 1: Define
Identify
\(\displaystyle y = \frac{x \ln x}{1 + \ln x}\)
Step 2: Differentiate
- Derivative Rule [Quotient Rule]: \(\displaystyle y' = \frac{(x \ln x)'(1 + \ln x) - (x \ln x)(1 + \ln x)'}{(1 + \ln x)^2}\)
- Derivative Rule [Product Rule]: \(\displaystyle y' = \frac{[(x)' \ln x + x(\ln x)'](1 + \ln x) - (x \ln x)(1 + \ln x)'}{(1 + \ln x)^2}\)
- Logarithmic Differentiation [Derivative Properties]: \(\displaystyle y' = \frac{[(x)' \ln x + 1](1 + \ln x) - \ln x}{(1 + \ln x)^2}\)
- Basic Power Rule: \(\displaystyle y' = \frac{\big( \ln (x) + 1 \big) (1 + \ln x) - \ln x}{(1 + \ln x)^2}\)
- Simplify: \(\displaystyle y' = \frac{(1 + \ln x)^2 - \ln x}{(1 + \ln x)^2}\)
- Rewrite: \(\displaystyle y' = \frac{(1 + \ln x)^2}{(1 + \ln x)^2} - \frac{\ln x}{(1 + \ln x)^2}\)
- Simplify: \(\displaystyle y' = 1 - \frac{\ln x}{(1 + \ln x)^2}\)
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Differentiation
