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If xy + 6ey = 6e,   find the value of y''   at the point where x = 0.?

If xy + 6ey = 6e, 

 find the value of y'' 

 at the point where

x = 0.

Update:

If

xy + 6e^y = 6e, find the value of y''

at the point where x = 0.

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This is the best answer for If xy + 6ey = 6e, find the value of y'' at the point where x = 0.?

xy + 6e^y = 6e passes through (0,1)

1st DI: x y' + y + y' 6e^y = 0 → y' = -y/(x + 6e^y)

2nd DI: x y'' + y' + y' + y'' 6e^y + (y')² 6e^y = 0

 → y'' = -(2y' + (y')² 6e^y)/(x + 6e^y)

 → y'' = -(2(-y/(x + 6e^y)) + (-y/(x + 6e^y))² 6e^y)/(x + 6e^y)

 → y''(0) = -(2(-1/(0 + 6e^1)) + (-1/(0 + 6e^1))² 6e^1)/(0 + 6e^1) → y''(0) = -(-2/(6e) + (1/(6e))/(6e) → y''(0) = -((-2+1)/(6e))/(6e) → y''(0) = 1/(36e²)

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Assuming you meant xy + 6e^y = 6e.

Use implicit differentiation (each side). What does that give?

Answer that and we will proceed.

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