1. State the problem.\nFind $\dfrac{dy}{dx}$ for the parametric equations $y=\dfrac{\theta-1}{\theta+1}$ and $x=\dfrac{\theta^2-1}{\theta^2+1}$.\n\n2. Use the parametric derivative rule (chain rule).\n$$\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$$\n\n3. Differentiate $y=\dfrac{\theta-1}{\theta+1}$.\nLet $u=\theta-1$ and $v=\theta+1$. Then $u'=1$ and $v'=1$.\n$$\frac{dy}{d\theta}=\frac{u'v-uv'}{v^2}=\frac{1(\theta+1)-(\theta-1)\cdot 1}{(\theta+1)^2}$$\n$$\frac{dy}{d\theta}=\frac{\theta+1-\theta+1}{(\theta+1)^2}=\frac{2}{(\theta+1)^2}$$\n\n4. Differentiate $x=\dfrac{\theta^2-1}{\theta^2+1}$.\nLet $p=\theta^2-1$ and $q=\theta^2+1$. Then $p'=2\theta$ and $q'=2\theta$.\n$$\frac{dx}{d\theta}=\frac{p'q-pq'}{q^2}=\frac{(2\theta)(\theta^2+1)-(\theta^2-1)(2\theta)}{(\theta^2+1)^2}$$\n$$\frac{dx}{d\theta}=\frac{2\theta\left[(\theta^2+1)-(\theta^2-1)\right]}{(\theta^2+1)^2}$$\n$$\frac{dx}{d\theta}=\frac{2\theta(2)}{(\theta^2+1)^2}=\frac{4\theta}{(\theta^2+1)^2}$$\n\n5. Divide to get $\dfrac{dy}{dx}$.\n$$\frac{dy}{dx}=\frac{\frac{2}{(\theta+1)^2}}{\frac{4\theta}{(\theta^2+1)^2}}$$\n$$\frac{dy}{dx}=\frac{2}{(\theta+1)^2}\cdot \frac{(\theta^2+1)^2}{4\theta}$$\n$$\frac{dy}{dx}=\frac{(\theta^2+1)^2}{\cancel{(\theta+1)^2}\cdot \cancel{2}\theta}\;\text{(combine constants and keep algebra steps)}$$\n$$\frac{dy}{dx}=\frac{2(\theta^2+1)^2}{4\theta(\theta+1)^2}$$\n$$\frac{dy}{dx}=\frac{(\theta^2+1)^2}{2\theta(\theta+1)^2}$$\n\n6. Final answer.\n$$\boxed{\frac{dy}{dx}=\frac{(\theta^2+1)^2}{2\theta(\theta+1)^2}}$$