x=θ+1θ and y=θ-1θ then dydx=

A
$\frac{x}{y}$
B
$\frac{y}{x}$
C
$\frac{- x}{y}$
D
$\frac{- y}{x}$

Solution:

Use the parametric differentiation

$\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}}$

So that

$\begin{array}{rcl} \frac{dy}{dx} & = & \frac{1 + \frac{1}{θ^{2}}}{1 - \frac{1}{θ^{2}}} \\ = & \frac{θ^{2} + 1}{θ^{2} - 1} \\ = & \frac{x}{y} \end{array}$

$x = θ + \frac{1}{θ} and y = θ - \frac{1}{θ} then \frac{dy}{dx} =$

Solution:

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