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If $t$ is a parameter and $x=t+\frac{1}{t}, y=t-\frac{1}{t}$, then $\frac{d^2 y}{d x^2}=$
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Verified Answer
The correct answer is:
$\frac{-4 t^3}{\left(t^2-1\right)^3}$
We have,
$$
\begin{aligned}
x & =t+\frac{1}{t}, y=t-\frac{1}{t} \\
\therefore \quad \frac{d x}{d t} & =1-\frac{1}{t^2}, \frac{d y}{d t}=1+\frac{1}{t^2} \\
\frac{d y}{d x} & =\frac{d y / d t}{d x / d t}=\frac{1+\frac{1}{t^2}}{1-\frac{1}{t^2}}=\frac{t^2+1}{t^2-1} \\
\Rightarrow \quad \frac{d^2 y}{d x^2} & =\frac{\left(t^2-1\right)(2 t)-\left(t^2+1\right)(2 t)}{\left(t^2-1\right)^2} \times \frac{d t}{d x} \\
\Rightarrow \quad \frac{d^2 y}{d x^2} & =\frac{2 t\left(t^2-1-t^2-1\right)}{\left(t^2-1\right)^2} \times \frac{t^2}{\left(t^2-1\right)} \\
\frac{d^2 y}{d x^2} & =\frac{-4 t^3}{\left(t^2-1\right)^3}
\end{aligned}
$$
$$
\begin{aligned}
x & =t+\frac{1}{t}, y=t-\frac{1}{t} \\
\therefore \quad \frac{d x}{d t} & =1-\frac{1}{t^2}, \frac{d y}{d t}=1+\frac{1}{t^2} \\
\frac{d y}{d x} & =\frac{d y / d t}{d x / d t}=\frac{1+\frac{1}{t^2}}{1-\frac{1}{t^2}}=\frac{t^2+1}{t^2-1} \\
\Rightarrow \quad \frac{d^2 y}{d x^2} & =\frac{\left(t^2-1\right)(2 t)-\left(t^2+1\right)(2 t)}{\left(t^2-1\right)^2} \times \frac{d t}{d x} \\
\Rightarrow \quad \frac{d^2 y}{d x^2} & =\frac{2 t\left(t^2-1-t^2-1\right)}{\left(t^2-1\right)^2} \times \frac{t^2}{\left(t^2-1\right)} \\
\frac{d^2 y}{d x^2} & =\frac{-4 t^3}{\left(t^2-1\right)^3}
\end{aligned}
$$
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