Answer:
[tex]\displaystyle f'(x) = \bigg( \frac{1}{2\sqrt{x}} - \sqrt{x} \bigg)e^\big{-x}[/tex]
General Formulas and Concepts:
Algebra I
Terms/Coefficients
Calculus
Differentiation
- Derivatives
- Derivative Notation
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 f(x) = \frac{\sqrt{x}}{e^x}[/tex]
Step 2: Differentiate
- Derivative Rule [Quotient Rule]: [tex]\displaystyle f'(x) = \frac{(\sqrt{x})'e^x - \sqrt{x}(e^x)'}{(e^x)^2}[/tex]
- Basic Power Rule: [tex]\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}(e^x)'}{(e^x)^2}[/tex]
- Exponential Differentiation: [tex]\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x}{(e^x)^2}[/tex]
- Simplify: [tex]\displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x}{e^{2x}}[/tex]
- Rewrite: [tex]\displaystyle f'(x) = \bigg( \frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x \bigg) e^{-2x}[/tex]
- Factor: [tex]\displaystyle f'(x) = \bigg( \frac{1}{2\sqrt{x}} - \sqrt{x} \bigg)e^\big{-x}[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Differentiation
