If y = frac{ax + b}{cx + d}, then 2 frac{dy}{ | vedclass

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(A) Given $y = \frac{ax + b}{cx + d}$Differentiating with respect to $x$, we get:$\frac{dy}{dx} = \frac{a(cx+d) - c(ax+b)}{(cx+d)^2} = \frac{ad-bc}{(c

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$3 \left( \frac{d^2y}{dx^2} \right)^2$

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(A) Given $y = \frac{ax + b}{cx + d}$Differentiating with respect to $x$, we get:$\frac{dy}{dx} = \frac{a(cx+d) - c(ax+b)}{(cx+d)^2} = \frac{ad-bc}{(cx+d)^2}$Differentiating again with respect to $x$:$\frac{d^2y}{dx^2} = \frac{-2c(ad-bc)}{(cx+d)^3}$Differentiating once more with respect to $x$:$\frac{d^3y}{dx^3} = \frac{6c^2(ad-bc)}{(cx+d)^4}$Now, $2 \frac{dy}{dx} \frac{d^3y}{dx^3} = 2 \left( \frac{ad-bc}{(cx+d)^2} \right) \left( \frac{6c^2(ad-bc)}{(cx+d)^4} \right) = \frac{12c^2(ad-bc)^2}{(cx+d)^6}$And $3 \left( \frac{d^2y}{dx^2} \right)^2 = 3 \left( \frac{-2c(ad-bc)}{(cx+d)^3} \right)^2 = \frac{12c^2(ad-bc)^2}{(cx+d)^6}$Therefore, $2 \frac{dy}{dx} \frac{d^3y}{dx^3} = 3 \left( \frac{d^2y}{dx^2} \right)^2$.

If $y = \frac{ax + b}{cx + d}$, then $2 \frac{dy}{dx} \frac{d^3y}{dx^3}$ is equal to

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