Find the derivative of y=x²−5x+4 from the First Principle.

Given $f(x) = x^2 - 5x + 4$.
Substitute $x+h$ into $f(x)$:

$$f(x+h) = (x+h)^2 - 5(x+h) + 4$$

Expand the terms:

$$f(x+h) = (x^2 + 2xh + h^2) - (5x + 5h) + 4$$ $$f(x+h) = x^2 + 2xh + h^2 - 5x - 5h + 4$$

$$f(x+h) - f(x) = (x^2 + 2xh + h^2 - 5x - 5h + 4) - (x^2 - 5x + 4)$$

Distribute the negative sign:

$$f(x+h) - f(x) = x^2 + 2xh + h^2 - 5x - 5h + 4 - x^2 + 5x - 4$$

Combine like terms:

$$f(x+h) - f(x) = 2xh + h^2 - 5h$$

$$\frac{f(x+h) - f(x)}{h} = \frac{2xh + h^2 - 5h}{h}$$

Factor out $h$ from the numerator:

$$\frac{f(x+h) - f(x)}{h} = \frac{h(2x + h - 5)}{h}$$

Cancel out $h$ (since $h \neq 0$ in the limit process):

$$\frac{f(x+h) - f(x)}{h} = 2x + h - 5$$

$$\frac{dy}{dx} = \lim_{h \to 0} (2x + h - 5)$$

Substitute $h=0$:

$$\frac{dy}{dx} = 2x + 0 - 5$$ $$\frac{dy}{dx} = 2x - 5$$

The derivative of $y = x^2 - 5x + 4$ from the first principle is:

$$2x - 5$$