How do you find (d^2y)/(dx^2)d2ydx2 for 5=x^2-2y^25=x2−2y2?
3 Answers
Oct 14, 2016
Deleted, because it was incorrect
I get
Explanation:
= ((1)(2y)-x(2(dy/dx)))/(2y)^2=(1)(2y)−x(2(dydx))(2y)2
= (2y-2x(x/(2y)))/(4y^2)=2y−2x(x2y)4y2
= (y-x(x/(2y)))/(2y^2)=y−x(x2y)2y2
= (y-x(x/(2y)))/(2y^2) * (2y)/(2y)=y−x(x2y)2y2⋅2y2y
= (2y^2-x^2)/(4y^3=2y2−x24y3
We started with
Oct 14, 2016