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

3 Answers
Oct 14, 2016

Deleted, because it was incorrect

Oct 14, 2016

I get #-5/(4y^3)#

Explanation:

#2y^2=x^2-5#

#4y dy/dx = 2x#

#dy/dx = x/(2y)#

#d/dx(dy/dx) = d/dx(x/(2y))#

# = ((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 #5=x^2-2y^2#, so we have

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

#(d^2y)/dx^2 = -5/(4y^3)#

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#