Answer:
\(\displaystyle \frac{dy}{dx} = e^x \bigg( \frac{1}{x} - \frac{1}{x^2} \bigg)\)
General Formulas and Concepts:
Calculus
Differentiation
- Derivatives
- Derivative Notation
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
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
\(\displaystyle y = \frac{e^x}{x}\)
Step 2: Differentiate
- Derivative Rule [Quotient Rule]: \(\displaystyle y' = \frac{(e^x)'x - e^x(x)'}{x^2}\)
- Exponential Differentiation: \(\displaystyle y' = \frac{e^xx - e^x(x)'}{x^2}\)
- Basic Power Rule: \(\displaystyle y' = \frac{e^xx - e^x}{x^2}\)
- Factor: \(\displaystyle y' = \frac{e^x(x - 1)}{x^2}\)
- Rewrite: \(\displaystyle y' = e^x \bigg( \frac{x}{x^2} - \frac{1}{x^2} \bigg)\)
- Simplify: \(\displaystyle y' = e^x \bigg( \frac{1}{x} - \frac{1}{x^2} \bigg)\)
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Differentiation
