Answer:
[tex]\displaystyle y' = 2x - 5[/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]
Explanation:
Step 1: Define
Identify
[tex]\displaystyle y = x^2 - 5x[/tex]
Step 2: Differentiate
- Substitute in function [Definition of a Derivative]: [tex]\displaystyle y' = \lim_{h \to 0} \frac{[(x + h)^2 - 5(x + h)] - (x^2 - 5x)}{h}[/tex]
- Expand: [tex]\displaystyle y' = \lim_{h \to 0} \frac{x^2 + 2hx + h^2 - 5x - 5h - x^2 + 5x}{h}[/tex]
- Combine like terms: [tex]\displaystyle y' = \lim_{h \to 0} \frac{2hx + h^2 - 5h}{h}[/tex]
- Factor: [tex]\displaystyle y' = \lim_{h \to 0} \frac{h(2x + h - 5)}{h}[/tex]
- Simplify: [tex]\displaystyle y' = \lim_{h \to 0} (2x + h - 5)[/tex]
- Evaluate limit [Limit Rule - Variable Direct Substitution]: [tex]\displaystyle y' = 2x + 0 - 5[/tex]
- Simplify: [tex]\displaystyle y' = 2x - 5[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Differentiation
