If y = frac{ax+b}{cx+d} and frac{dx}{dy} = fr | vedclass

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(B) Given $y = \frac{ax+b}{cx+d}$.Rearranging for $x$:$y(cx+d) = ax+b$$cxy + dy = ax + b$$x(cy - a) = b - dy$$x = \frac{dy-b}{a-cy}$Differentiating wi

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$(a-c)^2$

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(B) Given $y = \frac{ax+b}{cx+d}$.Rearranging for $x$:$y(cx+d) = ax+b$$cxy + dy = ax + b$$x(cy - a) = b - dy$$x = \frac{dy-b}{a-cy}$Differentiating with respect to $y$:$\frac{dx}{dy} = \frac{(a-cy)(d) - (dy-b)(-c)}{(a-cy)^2}$$\frac{dx}{dy} = \frac{ad - cdy + cdy - bc}{(a-cy)^2} = \frac{ad-bc}{c^2y^2 - 2acy + a^2}$Comparing this with $\frac{ad-bc}{Py^2+Qy+R}$,we get:$P = c^2, Q = -2ac, R = a^2$Therefore,$P+Q+R = c^2 - 2ac + a^2 = (a-c)^2$.

If $y = \frac{ax+b}{cx+d}$ and $\frac{dx}{dy} = \frac{ad-bc}{Py^2+Qy+R}$,then $P+Q+R =$

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