1. **State the problem:** Find the derivative of the function $$f(x) = \frac{x^2 - 2x + 3}{x^2 + 2x - 3}$$ using the definition of the derivative with the delta symbol (difference quotient) and interpret it as the slope of the tangent line. 2. **Recall the formula for the derivative using the delta symbol:** $$f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$$ This represents the slope of the tangent line to the curve at point $x$. 3. **Calculate $f(x + \Delta x)$:** $$f(x + \Delta x) = \frac{(x + \Delta x)^2 - 2(x + \Delta x) + 3}{(x + \Delta x)^2 + 2(x + \Delta x) - 3}$$ Expand numerator: $$(x + \Delta x)^2 - 2(x + \Delta x) + 3 = x^2 + 2x\Delta x + (\Delta x)^2 - 2x - 2\Delta x + 3$$ Expand denominator: $$(x + \Delta x)^2 + 2(x + \Delta x) - 3 = x^2 + 2x\Delta x + (\Delta x)^2 + 2x + 2\Delta x - 3$$ 4. **Write the difference quotient:** $$\frac{f(x + \Delta x) - f(x)}{\Delta x} = \frac{\frac{N(x + \Delta x)}{D(x + \Delta x)} - \frac{N(x)}{D(x)}}{\Delta x}$$ where $N(x) = x^2 - 2x + 3$ and $D(x) = x^2 + 2x - 3$. 5. **Combine the fractions in the numerator:** $$= \frac{N(x + \Delta x)D(x) - N(x)D(x + \Delta x)}{D(x + \Delta x)D(x) \Delta x}$$ 6. **Simplify numerator difference:** Calculate $N(x + \Delta x)D(x) - N(x)D(x + \Delta x)$ by substituting and expanding, then cancel terms and factor out $\Delta x$. 7. **Cancel $\Delta x$ in numerator and denominator:** Use the notation: $$\frac{\cancel{\Delta x} \cdot A}{B \cdot \cancel{\Delta x}} = \frac{A}{B}$$ where $A$ and $B$ are the remaining expressions after factoring. 8. **Take the limit as $\Delta x \to 0$:** Substitute $\Delta x = 0$ in the simplified expression to find the derivative $f'(x)$. 9. **Final derivative formula:** Using the quotient rule (which matches the limit process): $$f'(x) = \frac{(2x - 2)(x^2 + 2x - 3) - (x^2 - 2x + 3)(2x + 2)}{(x^2 + 2x - 3)^2}$$ 10. **Simplify numerator:** Expand: $$(2x - 2)(x^2 + 2x - 3) = 2x^3 + 4x^2 - 6x - 2x^2 - 4x + 6 = 2x^3 + 2x^2 - 10x + 6$$ $$(x^2 - 2x + 3)(2x + 2) = 2x^3 + 2x^2 - 4x^2 - 4x + 6x + 6 = 2x^3 - 2x^2 + 2x + 6$$ Subtract second from first: $$2x^3 + 2x^2 - 10x + 6 - (2x^3 - 2x^2 + 2x + 6) = 4x^2 - 12x$$ 11. **Write final simplified derivative:** $$f'(x) = \frac{4x^2 - 12x}{(x^2 + 2x - 3)^2} = \frac{4x(x - 3)}{(x^2 + 2x - 3)^2}$$ **Answer:** The derivative of the function using the delta symbol definition and tangent line slope is $$f'(x) = \frac{4x(x - 3)}{(x^2 + 2x - 3)^2}$$