The Best Answer for What? If x^2 + xy + y^3 = 1, find the value of y''' at the point where x = 1.?
When x = 1, what is y? We have:
1 + y + y^3 = 1
y + y^3 = 0
y(y^2 + 1) = 0
y = 0 ... (since y^2 + 1 > 0)
Thus we are interested in the point (x, y) = (1, 0). Implicitly differentiating with respect to x:
2x + y + xy' + 3y^2y' = 0
When (x, y) = (1, 0), we have:
2(1) + 0 + 1y' + 3(0)^2y' = 0
y' = -2
Implicitly differentiating again:
2 + y' + xy'' + y' + 6y(y')^2 + 3y^2y'' = 0
2 + 2y' + xy'' + 6y(y')^2 + 3y^2y'' = 0
When (x, y, y') = (1, 0, -2), we have:
2 + 2(-2) + 1y'' + 6(0)(-2)^2 + 3(0)^2y'' = 0
y'' = 2 ... (good!)
Implicitly differentiating again:
2y'' + xy''' + y'' + 6(y')^3 + 12yy'y'' + 6yy'y'' + 3y^2y''' = 0
3y'' + xy''' + 6(y')^3 + 18yy'y'' + 3y^2y''' = 0
When (x, y, y', y'') = (1, 0, -2, 2), we have:
3(2) + 1y''' + 6(-2)^3 + 18(0)(-2)(2) + 3(0)^2y''' = 0
y''' = 42
Hope that helps!