Question

If 3x^2 + y^2 = 7 then evaluate the second derivative of y with respect to x when x = 1 and y = 2. Round your answer to 2 decimal places.

Answer 1

3x^2+y^2 = 7
6x + 2yy' = 0
y' = -3x/y

using implicit differentiation again,
6x + 2yy' = 0
6 + 2y'^2 + 2yy" = 0
y" = -(6+2y'^2)/2y
= -(6+2(9x^2/y^2))/(2y)
= -3(3x^2+y^2)/y^3
= -21/y^3

or, using the quotient rule,
y' = -3x/y
y" = ((-3)y - (-3x)(-3x/y))/y^2
= -3(3x^2+y^2)/y^3
= -21/y^3