First slide

Derivatives

Question

If $x = t + \frac{1}{t}$ ,and, $y = t - \frac{1}{t}$ ,then $\frac{dy}{dx}$ ,is equal to

Moderate

A

$\frac{t^{2} + 1}{t^{2} - 1}$
B

$\frac{1 + t^{2}}{1 - t^{2}}$
C

$\frac{1 - t^{2}}{1 + t^{2}}$
D

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

Solution

Given that'
$x = t + \frac{1}{t} and y = t - \frac{1}{t}$
we have,
${(t + \frac{1}{t})}^{2} - {(t - \frac{1}{t})}^{2} = 4$
$\Rightarrow x^{2} - y^{2} = 4$
Now, differentiating w.r.t.x, we get
$2 x - 2 y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{x}{y} = \frac{t^{2} + 1}{t^{2} - 1}$

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