Question

How to find dy/dx and d^2y/dx^2 in terms of parameter?

`y=4/t , x=t-4/t`

Answer 1

`(d^2y)/dx^2 = (8t^3)/(t^2+4)^3`

Explanation:

From the parametric equations:

`{(x=t-4/t),(y=4/t):}`

we can get:

`x = t-y`

Differentiate both sides with respect to `t`

`dx/(dt) = 1- (dy)/(dt)`

and then using the chain rule to express `dy/dt`:

`dx/(dt) = 1- (dy)/dx dx/(dt)`

`dx/(dt)(1+dy/dx) = 1`

`(1+dy/dx) = 1/(dx/(dt))`

`dy/dx = -1+1/(dx/(dt))`

Now:

`dx/(dt) = 1+4/t^2 = (t^2+4)/t^2`

so:

`dy/dx = -1+t^2/(t^2+4) = = (-t^2-4+t^2)/(t^2+4)= -4/(t^2+4)`

Differentiate again with respect to `x`:

`(d^2y)/dx^2 =d/dx( -4/(t^2+4)) = (8t)/(t^2+4)^2 (dt)/(dx)`

and as:

`(dt)/(dx) = 1/(dx/(dt)) = t^2/(t^2+4)`

finally we have:

`(d^2y)/dx^2 = (8t)/(t^2+4)^2 t^2/(t^2+4)`

`(d^2y)/dx^2 = (8t^3)/(t^2+4)^3`