First and second derivative of a parametric

1. Sep 8, 2010

Exeneva

1. The problem statement, all variables and given/known data
$$x = t - e^{t}$$

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

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

2. Relevant equations
Derivative equations.

3. 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:
$$d^{2}y/dx$$ =$$- (e^{t} - e^{2t})^{-2} (e^{t} - 2e^{2t})$$

Last edited: Sep 9, 2010
2. Sep 8, 2010

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?

3. Sep 9, 2010

Exeneva

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

4. Sep 9, 2010

Char. Limit

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

5. Sep 10, 2010

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.

6. Sep 10, 2010

vela

Staff Emeritus
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.

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