# 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