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
Answer:
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!