# If xy + 3e^{y} = 3e, find the value of y'' at the point where x = 0.

**Solution:**

Given, xy + 3e^{y} = 3e

Taking derivative,

d(xy + 3e^{y})/dx = d/dx (3e)

d(xy)/dx + d(3e^{y})/dx = d/dx (3e)

On solving,

(1 + xdy/dx) + 3e dy/dx = 0

On grouping,

dy(x+3e)/dx = -1

dy/dx = -1/(x+3e)

Now, calculating second order derivative

y^{''} = d^{2}y/dx^{2}+3e))/dx

d^{2}y/dx^{2} = 1/(x+3e)^{2}

Y’’ at x = 0 is found to be

d^{2}y/dx^{2} = 1/(3e)^{2}

d^{2}y/dx^{2} = 1/9e^{2}

Therefore, the value of y” at x = 0 is 1/9 e^{2}.

## If xy + 3e^{y} = 3e, find the value of y'' at the point where x = 0.

**Summary:**

For xy + 3e^{y} = 3e, the value of y'' at the point where x = 0 is 1/9e^{2}.

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