Answer:
[tex]\displaystyle f'(x) = \frac{-3}{2\sqrt{2 - 3x}}[/tex]
General Formulas and Concepts:
Algebra I
Terms/Coefficients
Functions
Calculus
Limits
Limit Rule [Variable Direct Substitution]: [tex]\displaystyle \lim_{x \to c} x = c[/tex]
Differentiation
- Derivatives
- Derivative Notation
- Definition of a Derivative: [tex]\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}[/tex]
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle f(x) = \sqrt{2 - 3x}[/tex]
Step 2: Differentiate
- Substitute in function [Definition of a Derivative]: [tex]\displaystyle f'(x) = \lim_{h \to 0} \frac{\sqrt{2 - 3(x + h)} - \sqrt{2 - 3x}}{h}[/tex]
- Expand: [tex]\displaystyle f'(x) = \lim_{h \to 0} \frac{\sqrt{2 - 3x - 3h} - \sqrt{2 - 3x}}{h}[/tex]
- Rationalize: [tex]\displaystyle f'(x) = \lim_{h \to 0} \frac{\sqrt{2 - 3x - 3h} - \sqrt{2 - 3x}}{h} \cdot \frac{\sqrt{2 - 3x - 3h} + \sqrt{2 - 3x}}{\sqrt{2 - 3x - 3h} + \sqrt{2 - 3x}}[/tex]
- Simplify: [tex]\displaystyle f'(x) = \lim_{h \to 0} \frac{2 - 3x - 3h - (2 - 3x)}{h(\sqrt{2 - 3x - 3h} + \sqrt{2 - 3x})}[/tex]
- Expand: [tex]\displaystyle f'(x) = \lim_{h \to 0} \frac{2 - 3x - 3h - 2 + 3x}{h(\sqrt{2 - 3x - 3h} + \sqrt{2 - 3x})}[/tex]
- Combine like terms: [tex]\displaystyle f'(x) = \lim_{h \to 0} \frac{-3h}{h(\sqrt{2 - 3x - 3h} + \sqrt{2 - 3x})}[/tex]
- Simplify: [tex]\displaystyle f'(x) = \lim_{h \to 0} \frac{-3}{\sqrt{2 - 3x - 3h} + \sqrt{2 - 3x}}[/tex]
- Evaluate limit [Limit Rule - Variable Direct Substitution]: [tex]\displaystyle f'(x) = \frac{-3}{\sqrt{2 - 3(0) - 3h} + \sqrt{2 - 3x}}[/tex]
- Simplify: [tex]\displaystyle f'(x) = \frac{-3}{2\sqrt{2 - 3x}}[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Differentiation