Question:

Step 1: Simplify the expression for $y$ by combining terms progressively. We start by combining the first two terms: $$1 + \frac{c_1}{x-c_1} = \frac{x-c_1+c_1}{x-c_1} = \frac{x}{x-c_1}$$ Step 2: Add the third term to the result from Step 1. $$\frac{x}{x-c_1} + \frac{c_2 x}{(x-c_1)(x-c_2)} = \frac{x(x-c_2) + c_2 x}{(x-c_1)(x-c_2)} = \frac{x^2 - c_2 x + c_2 x}{(x-c_1)(x-c_2)} = \frac{x^2}{(x-c_1)(x-c_2)}$$ Step 3: Add the fourth term to the result from Step 2. $$\frac{x^2}{(x-c_1)(x-c_2)} + \frac{c_3 x^2}{(x-c_1)(x-c_2)(x-c_3)} = \frac{x^2(x-c_3) + c_3 x^2}{(x-c_1)(x-c_2)(x-c_3)} = \frac{x^3 - c_3 x^2 + c_3 x^2}{(x-c_1)(x-c_2)(x-c_3)} = \frac{x^3}{(x-c_1)(x-c_2)(x-c_3)}$$ Therefore: $$y = \frac{x^3}{(x-c_1)(x-c_2)(x-c_3)}$$ Step 4: Take the natural logarithm of both sides to simplify differentiation. $$\ln y = \ln x^3 - \ln(x-c_1) - \ln(x-c_2) - \ln(x-c_3)$$ $$\ln y = 3\ln x - \ln(x-c_1) - \ln(x-c_2) - \ln(x-c_3)$$ Step 5: Differentiate both sides with respect to $x$ using logarithmic differentiation. $$\frac{1}{y}\frac{dy}{dx} = \frac{3}{x} - \frac{1}{x-c_1} - \frac{1}{x-c_2} - \frac{1}{x-c_3}$$ Step 6: Factor out $\frac{1}{x}$ from the right-hand side. $$\frac{1}{y}\frac{dy}{dx} = \frac{1}{x}\left[3 - \frac{x}{x-c_1} - \frac{x}{x-c_2} - \frac{x}{x-c_3}\right]$$ Step 7: Simplify each fraction using the identity $\frac{x}{x-c_i} = 1 + \frac{c_i}{x-c_i}$. For each term: $$3 - \frac{x}{x-c_1} - \frac{x}{x-c_2} - \frac{x}{x-c_3} = 3 - \left(1 + \frac{c_1}{x-c_1}\right) - \left(1 + \frac{c_2}{x-c_2}\right) - \left(1 + \frac{c_3}{x-c_3}\right)$$ $$= 3 - 3 - \frac{c_1}{x-c_1} - \frac{c_2}{x-c_2} - \frac{c_3}{x-c_3}$$ $$= -\frac{c_1}{x-c_1} - \frac{c_2}{x-c_2} - \frac{c_3}{x-c_3}$$ $$= \frac{c_1}{c_1-x} + \frac{c_2}{c_2-x} + \frac{c_3}{c_3-x}$$ Step 8: Solve for $\frac{dy}{dx}$ by multiplying both sides by $y$. $$\frac{dy}{dx} = y \cdot \frac{1}{x}\left[\frac{c_1}{c_1-x} + \frac{c_2}{c_2-x} + \frac{c_3}{c_3-x}\right]$$ $$\frac{dy}{dx} = \frac{y}{x}\left[\frac{c_1}{c_1-x} + \frac{c_2}{c_2-x} + \frac{c_3}{c_3-x}\right]$$ **Final Answer:** The derivative is: $$\boxed{\frac{dy}{dx} = \frac{y}{x}\left[\frac{c_1}{c_1-x} + \frac{c_2}{c_2-x} + \frac{c_3}{c_3-x}\right]}$$ This matches **Option 4**.