1. **State the problem:** Find the derivative of the function $$f(x) = \frac{x^2 - 2x + 3}{x^2 + 2x - 3}$$ using the concept of tangent lines. 2. **Recall the formula:** For a function $$f(x) = \frac{u(x)}{v(x)}$$, the derivative is given by the quotient rule: $$ f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2} $$ where $$u(x) = x^2 - 2x + 3$$ and $$v(x) = x^2 + 2x - 3$$. 3. **Find the derivatives of numerator and denominator:** $$ u'(x) = \frac{d}{dx}(x^2 - 2x + 3) = 2x - 2$$ $$v'(x) = \frac{d}{dx}(x^2 + 2x - 3) = 2x + 2$$ 4. **Apply the quotient rule:** $$ f'(x) = \frac{(x^2 + 2x - 3)(2x - 2) - (x^2 - 2x + 3)(2x + 2)}{(x^2 + 2x - 3)^2} $$ 5. **Expand the terms:** $$ (x^2 + 2x - 3)(2x - 2) = 2x^3 + 4x^2 - 6x - 2x^2 - 4x + 6 = 2x^3 + 2x^2 - 10x + 6 $$ $$ (x^2 - 2x + 3)(2x + 2) = 2x^3 + 2x^2 - 4x^2 - 4x + 6x + 6 = 2x^3 - 2x^2 + 2x + 6 $$ 6. **Substitute back and simplify numerator:** $$ (2x^3 + 2x^2 - 10x + 6) - (2x^3 - 2x^2 + 2x + 6) = 2x^3 + 2x^2 - 10x + 6 - 2x^3 + 2x^2 - 2x - 6 $$ $$ = (2x^3 - 2x^3) + (2x^2 + 2x^2) + (-10x - 2x) + (6 - 6) = 0 + 4x^2 - 12x + 0 = 4x^2 - 12x $$ 7. **Write the derivative:** $$ f'(x) = \frac{4x^2 - 12x}{(x^2 + 2x - 3)^2} $$ 8. **Factor numerator:** $$ f'(x) = \frac{4x(x - 3)}{(x^2 + 2x - 3)^2} $$ **Final answer:** $$ f'(x) = \frac{4x(x - 3)}{(x^2 + 2x - 3)^2} $$ This derivative represents the slope of the tangent line to the curve at any point $$x$$.