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(A) Given $xy = 1$ and $x > 0$,we have $y = \frac{1}{x}$.Let $f(x) = x + y = x + \frac{1}{x}$.To find the minimum value,we differentiate $f(x)$ with r

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$2$

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(A) Given $xy = 1$ and $x > 0$,we have $y = \frac{1}{x}$.Let $f(x) = x + y = x + \frac{1}{x}$.To find the minimum value,we differentiate $f(x)$ with respect to $x$:$f'(x) = 1 - \frac{1}{x^2}$.Setting $f'(x) = 0$ for critical points:$1 - \frac{1}{x^2} = 0 \Rightarrow x^2 = 1$.Since $x > 0$,we take $x = 1$.Now,find the second derivative:$f''(x) = \frac{2}{x^3}$.At $x = 1$,$f''(1) = \frac{2}{1^3} = 2 > 0$.Since the second derivative is positive,$x = 1$ is a point of local minima.The minimum value is $f(1) = 1 + \frac{1}{1} = 2$.

$x$ and $y$ are two variables such that $x > 0$ and $xy = 1$. Then the minimum value of $x + y$ is

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