How do you find #(d^2y)/(dx^2)# for #5=x^2-2y^2#?
3 Answers
Oct 14, 2016
Deleted, because it was incorrect
I get
Explanation:
# = ((1)(2y)-x(2(dy/dx)))/(2y)^2#
# = (2y-2x(x/(2y)))/(4y^2)#
# = (y-x(x/(2y)))/(2y^2)#
# = (y-x(x/(2y)))/(2y^2) * (2y)/(2y)#
# = (2y^2-x^2)/(4y^3#
We started with
Oct 14, 2016