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