How do you find (d^2y)/(dx^2) given 2x^2-3y^2=4?

2 Answers
Oct 31, 2016

(d^2y)/(dx^2) = -8/(9y^3)

Explanation:

2x^2-3y^2=4

Differentiating implicitly we get;
4x-6ydy/dx=0
:. dy/dx=(4x)/(6y)
:. dy/dx=(2x)/(3y)

So, using the quotient rule
d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2
the second derivative is given by:

(d^2y)/(dx^2) = ((3y)(d/dx2x)-(2x)(d/dx3y))/(3y)^2
:. (d^2y)/(dx^2) = (6y-(2x)(dy/dxd/dy3y))/(9y^2)
:. (d^2y)/(dx^2) = (6y-(2x)(3dy/dx))/(9y^2)
:. (d^2y)/(dx^2) = (6y-6xdy/dx)/(9y^2)
:. (d^2y)/(dx^2) = (6y-6x((2x)/(3y)))/(9y^2)
:. (d^2y)/(dx^2) = (6y-4x^2/y)/(9y^2)
:. (d^2y)/(dx^2) = (6y^2-4x^2)/(9y^3)
:. (d^2y)/(dx^2) = 2(3y^2-2x^2)/(9y^3)

But 2x^2-3y^2=4 => 3y^2-2x^2=-4
:. (d^2y)/(dx^2) = 2(-4)/(9y^3)
:. (d^2y)/(dx^2) = -8/(9y^3)

Oct 31, 2016

color(red)((d^2y)/dx^2)=(2y-2x((dy)/dx))/(3y^2)

Explanation:

Computing the second implicit differentiation is determined by computing the first implicit derivative .

(2dx^2)/dx-(3dy^2)/dx=(d4)/dx
rArr2(2x)-3xx2yxx(dy)/dx=0
rArr4x-6yxx(dy)/dx=0
rArr-6yxx(dy)/dx=-4x
rArr(dy)/dx=(-4x)/(-6y)

rArrcolor(blue)((dy)/dx=(2x)/(3y))

color(red)((d^2y)/dx^2=??)

color(red)((d^2y)/dx^2)=((d(2x))/(dx)xx3y-3((dy)/dx)xx2x)/(3y)^2

color(red)((d^2y)/dx^2)=(2xx3y-6x((dy)/dx))/(3y)^2

color(red)((d^2y)/dx^2)=(6y-6x((dy)/dx))/(9y^2)

color(red)((d^2y)/dx^2)=(2y-2x((dy)/dx))/(3y^2)