Find `(d^2y)/(dx^2)` if `x^2-2y^2=3`

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The variables x and y are related by `x^2-2y^2=3` .

Taking the derivative of both the sides with respect to x,

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

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

The second derivative `(d^2y)/(dx^2) = (2y - 2*x*(dy/dx))/(4y^2)`

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

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

= `1/(2y) - x^2/(4*y^3) `

The required derivative `(d^2y)/(dx^2) = 1/(2y) - x^2/(4*y^3)`

pramodpandey's profile pic

pramodpandey | College Teacher | (Level 3) Valedictorian

Posted on

`` 

Differentiate with respect to x (implicitly)

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

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

`(dy)/(dx)=(1/2)xy^(-1)`       (ii)

differentate again w.r.t x

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

`=(1/2){y^(-1)-xy^(-2)((xy^(-1))/2)}`

`=(1/2){y^(-1)-(1/2)x^2y^(-3)}`

`=(1/(4y)){2-x^2y^(-2)}`

``

oldnick's profile pic

oldnick | (Level 1) Valedictorian

Posted on

`x^2-2y^2=3` 

expliciting respect y:

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

`y'=sqrt(2)/2x/sqrt(x^2-3)`

`y''=sqrt(2)/2(sqrt(x^2-3)-(x^2)/sqrt(x^2-3))/(x^2-3)=sqrt(2)/2(x^2-3-x^2)/(x^2-3)^(3/2)=-(3sqrt(2))/(2(x^2-3)^(3/2))`

Black line:  y(x)

Red line:   y'(x)

Blue line: y''(x)

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