1. **Problem statement:** Calculate the derivative of the function $$f(x) = \frac{2x + 4}{x - 2}$$ defined on the interval $$l = ]2 ; +\infty[,$$ and verify the derivative using an alternative expression of $$f(x)$$. 2. **Step 1: Calculate $$f'(x)$$ using the quotient rule.** The quotient rule states: $$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$$ where $$u(x) = 2x + 4$$ and $$v(x) = x - 2$$. Calculate derivatives: $$u'(x) = 2$$ $$v'(x) = 1$$ Apply the quotient rule: $$f'(x) = \frac{2(x - 2) - (2x + 4)(1)}{(x - 2)^2}$$ 3. **Step 2: Simplify the numerator:** $$2(x - 2) - (2x + 4) = 2x - 4 - 2x - 4 = -8$$ So, $$f'(x) = \frac{-8}{(x - 2)^2}$$ 4. **Step 3: Verify the alternative expression of $$f(x)$$.** Given: $$f(x) = 2 + \frac{8}{x - 2}$$ 5. **Step 4: Calculate $$f'(x)$$ using the alternative expression.** Derivative of constant 2 is 0. Derivative of $$\frac{8}{x - 2} = 8(x - 2)^{-1}$$ is: $$8 \times (-1)(x - 2)^{-2} = -\frac{8}{(x - 2)^2}$$ So, $$f'(x) = -\frac{8}{(x - 2)^2}$$ 6. **Step 5: Check consistency.** Both methods give the same derivative: $$f'(x) = -\frac{8}{(x - 2)^2}$$ 7. **Step 6: Calculate the derivative at $$x = 3$$.** $$f'(3) = -\frac{8}{(3 - 2)^2} = -\frac{8}{1^2} = -8$$ 8. **Step 7: Equation of the tangent line $$T$$ at $$x = 3$$.** The point on the curve is: $$f(3) = \frac{2(3) + 4}{3 - 2} = \frac{6 + 4}{1} = 10$$ The tangent line equation is: $$y = f'(3)(x - 3) + f(3) = -8(x - 3) + 10 = -8x + 24 + 10 = -8x + 34$$ **Final answer:** $$f'(x) = -\frac{8}{(x - 2)^2}$$ The tangent line at $$x=3$$ is: $$y = -8x + 34$$