How do you find (d^2y)/(dx^2)d2ydx2 for 5=x^2-2y^25=x22y2?

3 Answers
Oct 14, 2016

Deleted, because it was incorrect

Oct 14, 2016

I get -5/(4y^3)54y3

Explanation:

2y^2=x^2-52y2=x25

4y dy/dx = 2x4ydydx=2x

dy/dx = x/(2y)dydx=x2y

d/dx(dy/dx) = d/dx(x/(2y))ddx(dydx)=ddx(x2y)

= ((1)(2y)-x(2(dy/dx)))/(2y)^2=(1)(2y)x(2(dydx))(2y)2

= (2y-2x(x/(2y)))/(4y^2)=2y2x(x2y)4y2

= (y-x(x/(2y)))/(2y^2)=yx(x2y)2y2

= (y-x(x/(2y)))/(2y^2) * (2y)/(2y)=yx(x2y)2y22y2y

= (2y^2-x^2)/(4y^3=2y2x24y3

We started with 5=x^2-2y^25=x22y2, so we have

2y^2-x^2=-52y2x2=5, making the second derivative,

(d^2y)/dx^2 = -5/(4y^3)d2ydx2=54y3

Oct 14, 2016

y'' = -5/4 1/y^3

Explanation:

2y(x)^2-x^2+5=0->d/(dx)(2y(x)^2-x^2+5)=4y y'-2x=0

d/(dx)(2y y'-x)=2((y')^2+y y'')-1=0 so

y''=(1/2-(y')^2)/y but y'=1/2x/y so

y'' = (2y^2-x^2)/(4y^3) = -5/4 1/y^3