dw/dt for a function

Hi, I need to find dw/dt for w=xy + ln(xy), x=e^t, and y=e^(-t).  I guess I go ahead and add the derivatives of x and y, giving -->   ydx + 1/(xy) * ydx + xdy + 1/(xy) * xdy = 1 + 1 -1 -1 = 0 (???)
patricio26Asked:
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Infinity08Commented:
>> Well, dx/dt = e^t, and dy/dt = -e^(-t), no?

Right.

So when placing that in :

        ydx + 1/(xy) * ydx + xdy + 1/(xy) * xdy

you get :

        1 + 1 -1 -1 = 0

like you said :)
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Infinity08Commented:
>> ydx + 1/(xy) * ydx + xdy + 1/(xy) * xdy

ok.

Now, remember that x=e^t, and y=e^(-t).

What's dx/dt and dy/dt then ?
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patricio26Author Commented:
Well, dx/dt = e^t, and dy/dt = -e^(-t), no?
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patricio26Author Commented:
Great, thanks!
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phoffricCommented:
Here's a way to take less derivatives:

x(t) * y(t) = e^t * e^(-t) = e^(t-t) = e^(0) = 1

w(t) = xy + ln(xy) = 1 + ln 1 = 1 + 0 = 1

i.e., w(t) = 1
so, w'(t) = 0
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