If y = frac{ax + b}{cx + d}, then frac{dx}{dy | vedclass

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(B) Given $y = \frac{ax + b}{cx + d}$.To find $\frac{dx}{dy}$, we first express $x$ in terms of $y$:$y(cx + d) = ax + b$$cyx + yd = ax + b$$cyx - ax =

Vedclass · Accepted Answer

$\frac{ad - bc}{(a - cy)^2}$

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(B) Given $y = \frac{ax + b}{cx + d}$.To find $\frac{dx}{dy}$, we first express $x$ in terms of $y$:$y(cx + d) = ax + b$$cyx + yd = ax + b$$cyx - ax = b - yd$$x(cy - a) = b - yd$$x = \frac{b - yd}{cy - a} = \frac{yd - b}{a - cy}$.Now, differentiate $x$ with respect to $y$ using the quotient rule $\frac{d}{dy} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}$:$\frac{dx}{dy} = \frac{d(yd - b) \cdot (a - cy) - (yd - b) \cdot d(a - cy)}{dy} \cdot \frac{1}{(a - cy)^2}$$\frac{dx}{dy} = \frac{d(a - cy) - (yd - b)(-c)}{(a - cy)^2}$$\frac{dx}{dy} = \frac{ad - cdy + cdy - bc}{(a - cy)^2}$$\frac{dx}{dy} = \frac{ad - bc}{(a - cy)^2}$.

If $y = \frac{ax + b}{cx + d}$, then $\frac{dx}{dy} = $

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