Find dy/dx given x = (1+t³)/(t²-1), y = t/(t²-1)

Find $\dfrac{dy}{dx}$ given $x = \dfrac{1+t^{3}}{t^{2}-1}$, $y = \dfrac{t}{t^{2}-1}$.

Step 1: Differentiate $x$

Quotient rule with $u = 1+t^{3}$, $u' = 3t^{2}$, $v = t^{2}-1$, $v' = 2t$: \[x'_{t} = \frac{3t^{2}\bigl(t^{2}-1\bigr) - \bigl(1+t^{3}\bigr)2t}{\bigl(t^{2}-1\bigr)^{2}} = \frac{t\bigl(3t^{3}-3t-2-2t^{3}\bigr)}{\bigl(t^{2}-1\bigr)^{2}} = \frac{t\bigl(t^{3}-3t-2\bigr)}{\bigl(t^{2}-1\bigr)^{2}}.\]

Step 2: Factor the cubic

Since $t = -1$ is a root, $t+1$ divides it; dividing out shows the root is double: \[t^{3}-3t-2 = (t+1)\bigl(t^{2}-t-2\bigr) = (t+1)(t+1)(t-2) = (t+1)^{2}(t-2).\]

With $t^{2}-1 = (t-1)(t+1)$, the squared factor $(t+1)^{2}$ cancels: $x'_{t} = \frac{t(t+1)^{2}(t-2)}{(t-1)^{2}(t+1)^{2}} = \frac{t(t-2)}{(t-1)^{2}}$.

Step 3: Differentiate $y$

\[ y'_{t} = \frac{1\cdot\bigl(t^{2}-1\bigr) - t\cdot 2t}{\bigl(t^{2}-1\bigr)^{2}} = \frac{-\bigl(t^{2}+1\bigr)}{(t-1)^{2}(t+1)^{2}}. \]

Step 4: Form the ratio

The factors $(t-1)^{2}$ cancel between numerator and denominator: \[\frac{dy}{dx} = \frac{-\bigl(t^{2}+1\bigr)}{(t-1)^{2}(t+1)^{2}}\cdot\frac{(t-1)^{2}}{t(t-2)} = -\frac{t^{2}+1}{t(t-2)(t+1)^{2}}.\]

Step 5: Verification

\[t = 3:\quad x = \frac{28}{8} = 3.5,\ y = \frac{3}{8} = 0.375\]

\[\frac{dy}{dx} = -\frac{10}{3\cdot 1\cdot 16} = -\frac{10}{48} \approx -0.20833\]

\[\text{Numerically, } \frac{y(3.001)-y(2.999)}{x(3.001)-x(2.999)} \approx -0.20833 \;\;\checkmark\]