Find the equation of tangent line. (1,2) on the curve xy=2 .Find the equation of tangent line on this curve at this point.

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You should remember the formula of equation of tangent line to the curve `xy=2` , at the point `(1,2)`  such that:

`y - 2 = f'(1)(x - 1)`

You need to find the derivative of the function `f(x) = y = 2/x` , using the quotient rule such that:

`f'(x) =...

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You should remember the formula of equation of tangent line to the curve `xy=2` , at the point `(1,2)`  such that:

`y - 2 = f'(1)(x - 1)`

You need to find the derivative of the function `f(x) = y = 2/x` , using the quotient rule such that:

`f'(x) = (2'*x - 2*x')/x^2 => f'(x) = (0 - 2*1)/x^2 => f'(x) = -2/x^2`

You need to evaluate f'(1) such that:

`f'(1) = -2/1 = -2`

You need to substitute -2 for f'(1) in equation of tangent line such that:

`y - 2 = -2(x - 1) => y = -2x + 2 + 2 => y = -2x + 4`

Hence, evaluating the equation of tangent line to the given curve, at the point `(1,2), `  yields `y = -2x + 4` .

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