Question

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

Answer 1

`(d^2y)/(dx^2) =-1/(sqrt5(5-2x)^(3/2))`

Explanation:

It would be possible to explicit `y(x)= +-sqrt((2x-5)/5)`, but we will use the procedure for implicit differentiation.

Differentiate the equation:

`2x-5y^2=5`

term by term with respect to `x`, considering that based on the chain rule:

`(dg(y))/(dx) = (dg(y))/(dy)*(dy)/(dx)`

So, we have:

`2-10y(dy)/(dx)= 0`

Solving for `(dy)/(dx)`:

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

Differentiating again:

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

Now from the original equation we have:

`5y^2=5-2x`

and we can use this to substitute `y` in the expression of the second derivative:

`(d^2y)/(dx^2) = -1/(25y^3) = -1/sqrt(5) * 1/(5y^2)^(3/2)=-1/(sqrt5(5-2x)^(3/2))`