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