First and second derivative of a parametric

[Exeneva]

Homework Statement

$$x = t - e^{t}$$

$$y = t + e^{-t}$$

Find dy/dx and $$d^{2}y/dx.$$

Homework Equations

Derivative equations.

The Attempt at a Solution

dy/dt = $$1 - e^{-t}$$

dx/dt = $$1 - e^{t}$$

The dy/dx I came up with is:
dy/dx = $$(1 - e^{-t}) / (1 - e^{t})$$

Second derivative I came up with is:
$$d^{2}y/dx$$ =$$- (e^{t} - e^{2t})^{-2} (e^{t} - 2e^{2t})$$

 

[Dick]

Assuming you meant y=t+e^(-t) the first derivative looks ok. The second derivative doesn't. Mind showing us how you got it?

 

[Exeneva]

I think I tried to do a chain rule but screwed up somewhere.

 

[Char. Limit]

It's rather hard to find $\frac{d^2 y}{dx}$. You should be looking for $\frac{d^2 y}{d x^2}$.

 

[Exeneva]

Sorry, that's what I meant, the second derivative (just forgot the squared on the bottom).

My main issue is I'm not sure how I was supposed to have found the second derivative. I tried to do a reverse chain rule but I am pretty sure I did it wrong.

 

[vela]

I think I tried to do a chain rule but screwed up somewhere.

Sorry, that's what I meant, the second derivative (just forgot the squared on the bottom).

My main issue is I'm not sure how I was supposed to have found the second derivative. I tried to do a reverse chain rule but I am pretty sure I did it wrong.

It would help if you showed us your actual work. Just saying "I tried this and it didn't work" isn't very helpful.

By the way, the first derivative simplifies quite a bit, down to dy/dx=-e^-t.