Question

Find the derivative.

Answer 1

`\displaystyle f'(x) = \bigg( (1)/(2√(x)) - √(x) \bigg)e^\big{-x}`

General Formulas and Concepts:

Algebra I

Terms/Coefficients

• Expanding/Factoring

Calculus

Differentiation

• Derivatives
• Derivative Notation

Basic Power Rule:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Quotient Rule]:
`\displaystyle (d)/(dx) [(f(x))/(g(x)) ]=(g(x)f'(x)-g'(x)f(x))/(g^2(x))`

Explanation:

Step 1: Define

Identify

`\displaystyle f(x) = (√(x))/(e^x)`

Step 2: Differentiate

1. Derivative Rule [Quotient Rule]:
   `\displaystyle f'(x) = ((√(x))'e^x - √(x)(e^x)')/((e^x)^2)`
2. Basic Power Rule:
   `\displaystyle f'(x) = ((e^x)/(2√(x)) - √(x)(e^x)')/((e^x)^2)`
3. Exponential Differentiation:
   `\displaystyle f'(x) = ((e^x)/(2√(x)) - √(x)e^x)/((e^x)^2)`
4. Simplify:
   `\displaystyle f'(x) = ((e^x)/(2√(x)) - √(x)e^x)/(e^(2x))`
5. Rewrite:
   `\displaystyle f'(x) = \bigg( (e^x)/(2√(x)) - √(x)e^x \bigg) e^(-2x)`
6. Factor:
   `\displaystyle f'(x) = \bigg( (1)/(2√(x)) - √(x) \bigg)e^\big{-x}`

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation