Answer:
[tex]\displaystyle y' = \frac{1}{2\sqrt{x}}[/tex]
General Formulas and Concepts:
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 y = \sqrt{x}[/tex]
Step 2: Differentiate
- Substitute in function [Definition of a Derivative]: [tex]\displaystyle y' = \lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}[/tex]
- Rationalize: [tex]\displaystyle y' = \lim_{h \to 0} \frac{\sqrt{x + h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}}[/tex]
- Simplify: [tex]\displaystyle y' = \lim_{h \to 0} \frac{x + h - x}{h(\sqrt{x + h} + \sqrt{x})}[/tex]
- Simplify: [tex]\displaystyle y' = \lim_{h \to 0} \frac{1}{\sqrt{x + h} + \sqrt{x}}[/tex]
- Evaluate limit [Limit Rule - Variable Direct Substitution]: [tex]\displaystyle y' = \frac{1}{\sqrt{x + 0} + \sqrt{x}}[/tex]
- Simplify: [tex]\displaystyle y' = \frac{1}{2\sqrt{x}}[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Differentiation
