# find (d^2y)/(dx^2) if x^2-2y^2=3 (d^2y)/(dx^2)=

pramodpandey | Student

Given   `x^2-2y^2=3`     (i)

differentiate with respect to x (implicitly).

`2x-4y(dy)/(dx)=0`

`-4y(dy)/(dx)=-2x`

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

`(dy)/(dx)=(1/2)xy^(-1)`       (ii)

differentiate (ii) w.r.t. x ,

`(d^2y)/(dx^2)=(1/2)(y^(-1)+x(-1)y^(-2)(dy)/(dx))`

Substitute `(dy)/(dx)`from (ii) ,we have

`(d^2y)/(dx^2)=(1/2)(y^(-1)-xy^(-2)(1/2)xy^(-1))`

`` `=(1/2)(y^(-1)-(x^2y^(-3))/2)`

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

`=-3/(4y^3)`       ,   using(i)

oldnick | Student

first we have to  set  funtion in form  y = f(x)

`y=` `sqrt(2)/2` `sqrt(x^2 -3)`

now we have composed function derivative:

`d/dx`  `u[v(x)]= (du)/(dv)v'(x) `

` v(x) = x^2-3 and u(x)= sqrt[v(x)]`

`dy/dx=sqrt(2)/2 [2x ]/[2sqrt(x^2 - 3)]`

`semplifing:`

`dy/dx=sqrt(2)/2 (x)/sqrt(x^2 -3)`

``

`(d^2y)/(dx^2) = sqrt(2)/2 [sqrt(x^2 -3)-(x^2)/sqrt(x^2-3)]/[x^2-3)`

`(d^2y)/(dx^2)= sqrt(2)/2 [-3]/[sqrt(x^2 -3) (x^2 -3)]`

`(d^2y)/(dx^2)=-3sqrt(2)/2 1/[(x^2-3)^(3/2)]`

``