Question

How do you find the second derivative of `2x^2-y^2=1`?

Answer 1

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

Explanation:

We have
`2x^2-y^2=1`.........equation 1
Differentiating with respect to `x` we get
`4x-2y (dy)/(dx) =0`
`=>(dy)/(dx)=2x/y`
Again differentiating with respect to `x` we get
`(d^2y)/dx^2=2 ((y-x (dy)/(dx)))/y^2`
Now putting the value of `(dy)/(dx)` we get
`(d^2y)/dx^2=2( (y^2-2x^2))/y^3`
`:.(d^2y)/dx^2=-2/y^3` (using equation 1)