prove that the curve 4y-x^2-8=0 is increasing in (0,4)?

4y - x² - 8 = 0
(d/dx)(4y - x² - 8) = 0
4y' - 2x = 0
4y' = 2x
y' = x/2

y" = (d/dx)(x/2)
y" = 1/2

At (0, 4) :
y' = 0
y" = 1/2 > 0

Hence, the curve 4y - x² - 8 = 0 is increasing at (0, 4).

　
4y − x² − 8 = 0
4y' − 2x = 0
4y' = 2x
y' = x/2

For x on interval (0, 4), y' = x/2 > 0, so y is increasing on this interval.