Question Statement

Find $\frac{dy}{dx}$ given the parametric equations:

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

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Background and Explanation

This problem involves parametric differentiation, where both $x$ and $y$ are defined in terms of a parameter $t$. Since $y$ is not explicitly expressed as a function of $x$, we use the chain rule to relate the derivatives with respect to $t$.

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Solution

Step 1: Differentiate $x$ with respect to $t$

First, rewrite $x$ using negative exponents to apply the power rule easily:

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

Differentiate term by term using the power rule $\frac{d}{dt}(t^n)=n{t}^{n-1}$:

$\frac{dx}{dt}=\frac{d}{dt}(t^2)+\frac{d}{dt}(t^{-2})=2t+(-2)t^{-3}$

Simplify the expression:

$\frac{dx}{dt}=2t-2t^{-3}=2t-\frac{2}{t^3}$

Combine over a common denominator $t^3$:

$\frac{dx}{dt}=\frac{2t^4-2}{t^3}$

Take the reciprocal to find $\frac{dt}{dx}$ (needed for the chain rule):

$\frac{dt}{dx}=\frac{t^3}{2t^4-2}$

Step 2: Differentiate $y$ with respect to $t$

Rewrite $y$ using negative exponents:

$y=t-\frac{1}{t}=t-t^{-1}$

Differentiate term by term:

$\frac{dy}{dt}=\frac{d}{dt}(t)-\frac{d}{dt}(t^{-1})=1-(-1)t^{-2}$

Simplify:

$\frac{dy}{dt}=1+t^{-2}=1+\frac{1}{t^2}$

Combine over a common denominator $t^2$:

$\frac{dy}{dt}=\frac{t^2+1}{t^2}$

Step 3: Apply the Chain Rule

Using the chain rule for parametric equations:

$\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}$

Substitute the expressions found above:

$\frac{dy}{dx}=\frac{t^2+1}{t^2}\times \frac{t^3}{2t^4-2}$

Multiply the numerators and denominators:

$=\frac{(t^2+1)\times t^3}{t^2\times 2(t^4-1)}$

Cancel $t^2$ from numerator and denominator:

$=\frac{(t^2+1)\cdot t}{2(t^4-1)}$

Therefore, the final answer is:

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

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Key Formulas or Methods Used

• Chain Rule for Parametric Equations: $\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}$ (or equivalently $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$)
• Power Rule: $\frac{d}{dt}(t^n)=n{t}^{n-1}$ for any real number $n$
• Negative Exponents: $\frac{1}{t^n}=t^{-n}$ (used to rewrite terms before differentiating)
• Reciprocal Rule: $\frac{dt}{dx}=\frac{1}{\frac{dx}{dt}}$

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Summary of Steps

1. Rewrite $x$ and $y$ using negative exponents ($t^{-n}$ format) to prepare for differentiation
2. Differentiate $x$ with respect to $t$ to get $\frac{dx}{dt}$, then take the reciprocal to find $\frac{dt}{dx}$
3. Differentiate $y$ with respect to $t$ to get $\frac{dy}{dt}$
4. Apply the chain rule by multiplying: $\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}$
5. Simplify the resulting algebraic expression by factoring and canceling common terms