If y = frac{x^2}{(x - 1)(x - 2)(x - 3)} + fra | vedclass

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(C) First,simplify the expression for $y$ by finding a common denominator:$y = \frac{x^2 + 2x(x - 1) + 3(x - 1)(x - 2) + (x - 1)(x - 2)(x - 3)}{(x - 1

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$\frac{1}{1 - x} + \frac{2}{2 - x} + \frac{3}{3 - x}$

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(C) First,simplify the expression for $y$ by finding a common denominator:$y = \frac{x^2 + 2x(x - 1) + 3(x - 1)(x - 2) + (x - 1)(x - 2)(x - 3)}{(x - 1)(x - 2)(x - 3)}$$y = \frac{x^2 + 2x^2 - 2x + 3(x^2 - 3x + 2) + (x^2 - 3x + 2)(x - 3)}{(x - 1)(x - 2)(x - 3)}$$y = \frac{3x^2 - 2x + 3x^2 - 9x + 6 + x^3 - 3x^2 - 3x^2 + 9x + 2x - 6}{(x - 1)(x - 2)(x - 3)}$$y = \frac{x^3}{(x - 1)(x - 2)(x - 3)}$Taking the natural logarithm on both sides:$\ln y = 3 \ln x - \ln(x - 1) - \ln(x - 2) - \ln(x - 3)$Differentiating with respect to $x$:$\frac{y'}{y} = \frac{3}{x} - \frac{1}{x - 1} - \frac{1}{x - 2} - \frac{1}{x - 3}$Multiply by $x$:$\frac{xy'}{y} = 3 - \frac{x}{x - 1} - \frac{x}{x - 2} - \frac{x}{x - 3}$$= (1 - \frac{x}{x - 1}) + (1 - \frac{x}{x - 2}) + (1 - \frac{x}{x - 3})$$= \frac{x - 1 - x}{x - 1} + \frac{x - 2 - x}{x - 2} + \frac{x - 3 - x}{x - 3}$$= \frac{-1}{x - 1} + \frac{-2}{x - 2} + \frac{-3}{x - 3}$$= \frac{1}{1 - x} + \frac{2}{2 - x} + \frac{3}{3 - x}$

If $y = \frac{x^2}{(x - 1)(x - 2)(x - 3)} + \frac{2x}{(x - 2)(x - 3)} + \frac{3}{x - 3} + 1$,then $\frac{xy'}{y}$ is equal to (where $y' = \frac{dy}{dx}$):

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