Have i done this parametric differentiation right?

[bill nye scienceguy!]

y=t+cost x=t+sint

dy/dt=1-sint dx/dt=1+cost

dy/dx= (dy/dt).(dt/dx)

= (1-sint).1/(1+cost) = (1-sint)/(1+cost)
= 1-tant

and how do i get from there to the second order differential?

 

[VietDao29]

y=t+cost x=t+sint

dy/dt=1-sint dx/dt=1+cost

dy/dx= (dy/dt).(dt/dx)

= (1-sint).1/(1+cost) = (1-sint)/(1+cost)
= 1-tant

and how do i get from there to the second order differential?

How did you go from:
\frac{dy}{dx} = \frac{1 - \sin t}{1 + \cos t} \quad \mbox{to} \quad = 1 - \tan t?
Just stop at \frac{dy}{dx} = \frac{1 - \sin t}{1 + \cos t}. It's okay, you don't have to simplify it any further.
To find:
\frac{d ^ 2 y}{dx ^ 2}, we use the Chain rule, and the derivative of inverse function:
\frac{dk}{du} = \frac{1}{\frac{du}{dk}}
\frac{d ^ 2 y}{dx ^ 2} = \frac{d \left( \frac{dy}{dx} \right)}{dx} = \frac{d \left( \frac{dy}{dx} \right)}{dt} \times \frac{dt}{dx} = \frac{d \left( \frac{dy}{dx} \right)}{dt} \times \frac{1}{\frac{dx}{dt}}
Can you go from here? :)

 

[Rasine]

the second order derivitve is just the derivitive of dy/dx all divided by dx/dt