Question

How do you find `(d^2y)/(dx^2)` for `2x-5y^2=3`?

Answer 1

`(d^2y)/dx^2=-1/(25y^3)`

Explanation:

`2x-5y^2=3 =>(d(2x))/dx-(d(5y^2))/dx=(d(3))/dx=>`

`2-10ydy/dx=0=>1-5ydy/dx=0 ..(1)`

`(d(1))/dx-(d(5ydy/dx))/dx=0=>-(d(5ydy/dx))/dx=0=>`

`(d(5y))/dxdy/dx+5y(d(dy/dx))/dx=0=>`

`5(dy/dx)^2+5y(d^2y)/dx^2=0`

Let's substitude from the equation `(1)` that `dy/dx=1/(5y)`

So we get :

`5(1/(5y))^2+5y(d^2y)/dx^2=0=>1/(5y^2)+5y(d^2y)/dx^2=0=>`

`5y(d^2y)/dx^2=-1/(5y^2)=>(d^2y)/dx^2=-1/(25y^3)`

Answer 2

`((d^2y)/(dx^2))^2 = 1/5 1/(2x-3)^3`

Explanation:

Differentiate both sides of the equation with respect to `x`:

`d/dx (2x-5y^2) = 0`

`2-10y dy/dx = 0`

`10y dy/dx = 2`

`(1)" "dy/dx =1/(5y)`

and differentiating again:

`(d^2y)/(dx^2) = -1/(5y^2)dy/dx`

substitute now `dy/dx` from `(1)`:

`(2) " " (d^2y)/(dx^2) = -1/(25y^3)`

We can also have an implicit equation for `(d^2y)/(dx^2)` by squaring both sides of `(2)`:

`((d^2y)/(dx^2))^2 = 1/(5^4y^6) = 1/5^4 (1/y^2)^3`

and substituting `y^2 = (2x-3)/5` from the original equation:

`((d^2y)/(dx^2))^2 = 1/5 1/(2x-3)^3`

Function:

graph{y^2 = (2x-3)/5 [-10, 10, -5, 5]}

Derivative:

graph{y^2 = 1/5 1/(2x-3)^3 [-10, 10, -5, 5]}