Question

Calculate y' and y'' at the point (1,1) on the curve 5 xy^2+9 y-14 = 0

I got the first part (-5/19) but I can't get the answer for the second

Answer 1

first derivative:

5x y^2 + 9y - 14 = 0
5x(2y)dy/dx + 5y^2 + 9dy/dx = 0
dy/dx(10xy + 9) = -5y^2
dy/dx = -5y^2/(10xy+9)
at (1,1)
dy/dx
= y'
= -5/19 , you had that

from 5x(2y)dy/dx + 5y^2 + 9dy/dx = 0
10xy y' + 5y^2 + 9 y' = 0
recall the rule for derivatives of triple products
d(uvw) = uvw' + uv' w + u' vw

10(1)(y)(y') + 10x(y')(y') + 10xy(y'') + 10y y' + 9 y'' = 0
we can replace y' with -5/19

so .....
10(1)(1)(-5/19) + 10(1)(-5/19)(-5/19) + 10(1)(1)y'' + 10(1)(-5/19) + 9 y'' = 0

I will leave the arithmetic up to you

Answer 2

or, differentiating y' directly,

y" = 50y^3(-5xy^2+20xy+18)/(10xy+9)^3
y"(1,1) = 50(-5+20+18)/19^3 = 1650/6859

which agrees with Reiny's result above!!!