Question Statement

Given the parametric equations:

$x=\frac{{\theta }^2-1}{{\theta }^2+1}\text{and}y=\frac{\theta -1}{\theta +1}$

Find $\frac{dy}{dx}$.

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Background and Explanation

This problem involves parametric differentiation, where both $x$ and $y$ are defined in terms of a third variable (parameter) $\theta$. To find $\frac{dy}{dx}$, we calculate the derivatives of $x$ and $y$ with respect to $\theta$ separately, then apply the chain rule.

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Solution

Step 1: Find $\frac{dx}{d\theta }$

Given:

$x=\frac{{\theta }^2-1}{{\theta }^2+1}$

Differentiate with respect to $\theta$ using the quotient rule $\frac{d}{d\theta }\left(\frac{u}{v}\right)=\frac{v\frac{du}{d\theta }-u\frac{dv}{d\theta }}{v^2}$:

$\begin{array}{rl}\frac{dx}{d\theta }& =\frac{({\theta }^2+1)\frac{d}{d\theta }({\theta }^2-1)-({\theta }^2-1)\frac{d}{d\theta }({\theta }^2+1)}{({\theta }^2+1{)}^2}\\ & =\frac{({\theta }^2+1)(2\theta )-({\theta }^2-1)(2\theta )}{({\theta }^2+1{)}^2}\\ & =\frac{2{\theta }^3+2\theta -2{\theta }^3+2\theta }{({\theta }^2+1{)}^2}\\ & =\frac{4\theta }{({\theta }^2+1{)}^2}\end{array}$

Step 2: Find $\frac{dy}{d\theta }$

Given:

$y=\frac{\theta -1}{\theta +1}$

Differentiate with respect to $\theta$ using the quotient rule:

$\begin{array}{rl}\frac{dy}{d\theta }& =\frac{(\theta +1)\frac{d}{d\theta }(\theta -1)-(\theta -1)\frac{d}{d\theta }(\theta +1)}{(\theta +1{)}^2}\\ & =\frac{(\theta +1)(1)-(\theta -1)(1)}{(\theta +1{)}^2}\\ & =\frac{\theta +1-\theta +1}{(\theta +1{)}^2}\\ & =\frac{2}{(\theta +1{)}^2}\end{array}$

Step 3: Apply the Chain Rule

By the chain rule for parametric equations:

$\frac{dy}{dx}=\frac{dy}{d\theta }\times \frac{d\theta }{dx}=\frac{dy/d\theta }{dx/d\theta }$

Substitute the derivatives found above:

$\begin{array}{rl}\frac{dy}{dx}& =\frac{\frac{2}{(\theta +1{)}^2}}{\frac{4\theta }{({\theta }^2+1{)}^2}}\\ & =\frac{2}{(\theta +1{)}^2}\times \frac{({\theta }^2+1{)}^2}{4\theta }\\ & =\frac{({\theta }^2+1{)}^2}{2\theta (\theta +1{)}^2}\end{array}$

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Key Formulas or Methods Used

• Quotient Rule: $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$
• Chain Rule (Parametric): $\frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }$

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Summary of Steps

1. Differentiate $x$: Apply the quotient rule to $x=\frac{{\theta }^2-1}{{\theta }^2+1}$ to obtain $\frac{dx}{d\theta }=\frac{4\theta }{({\theta }^2+1{)}^2}$
2. Differentiate $y$: Apply the quotient rule to $y=\frac{\theta -1}{\theta +1}$ to obtain $\frac{dy}{d\theta }=\frac{2}{(\theta +1{)}^2}$
3. Apply chain rule: Divide $\frac{dy}{d\theta }$ by $\frac{dx}{d\theta }$ (or multiply by $\frac{d\theta }{dx}$)
4. Simplify: Cancel common factors to get the final result $\frac{({\theta }^2+1{)}^2}{2\theta (\theta +1{)}^2}$