Q.If $x = \frac{4t}{1 + t^2}, \ y = 3\left(\frac{1 - t^2}{1 + t^2}\right)$ then show that $\frac{dy}{dx} = \frac{-9x}{4y}$.

$\dfrac{dx}{dt}=\dfrac{4(1-t^2)}{(1+t^2)^2}$, $\dfrac{dy}{dt}=\dfrac{-12t}{(1+t^2)^2}$, so $\dfrac{dy}{dx}=\dfrac{-3t}{1-t^2}$; and $\dfrac{-9x}{4y}=\dfrac{-9\cdot 4t/(1+t^2)}{4\cdot3(1-t^2)/(1+t^2)}=\dfrac{-3t}{1-t^2}$ — the two agree.

Step 1 — differentiate $x=\dfrac{4t}{1+t^2}$ by the quotient rule:

$\frac{dx}{dt}=\frac{4(1+t^2)-4t(2t)}{(1+t^2)^2}=\frac{4(1-t^2)}{(1+t^2)^2}.$

Step 2 — differentiate $y=3\cdot\dfrac{1-t^2}{1+t^2}$:

$\frac{dy}{dt}=3\cdot\frac{(-2t)(1+t^2)-(1-t^2)(2t)}{(1+t^2)^2}=3\cdot\frac{-2t\big[(1+t^2)+(1-t^2)\big]}{(1+t^2)^2}=\frac{-12t}{(1+t^2)^2}.$

Step 3 — form $\dfrac{dy}{dx}$:

$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{-12t/(1+t^2)^2}{4(1-t^2)/(1+t^2)^2}=\frac{-12t}{4(1-t^2)}=\frac{-3t}{1-t^2}.$

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