# [SOLVED] *THIRD prime? What? If x^2 + xy + y^3 = 1, find the value of y''' at the point where x = 1.?

What? If x^2 + xy + y^3 = 1, find the value of y''' at the point where x = 1.?

If

x2 + xy + y3 = 1,

find the value of

y'''

at the point where

x = 1.

:(

I think second prime is 2 when x=1

...

stuck after that

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## 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!

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