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Derivatives of parametric curves: http://screencast.com/t/x5c5hBqr

Mathematics
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|dw:1351711571629:dw|
ok, d2y/dx^2 = d/dx (-cot t) , right ? d/dx (-cot t) = d/dt(-cot t) * dt/dx and u know dt/dx from dx/dt

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Other answers:

By what parameter you want to differentiate cot?
i think x, isn't it ?
|dw:1351712169898:dw|
u got what i did ?
when t = pi/4
I'm still confused how you got d/dt(-cot t) * dt/dx
|dw:1351712487011:dw|
\(\huge \frac{d}{dx}f(t)=\frac{d}{dt}f(t) \frac{dt}{dx}\)
oohhh.. okay... so we can write:|dw:1351712793765:dw|
|dw:1351712929460:dw|
seems correct, now u can put t=pi/4
|dw:1351713083407:dw|
yup.
hmmmm... okay thanks so much for the help... you're the best!
not really... welcome ^_^
i am still a little bit confused how this works: d/dx (-cot t) = d/dt(-cot t) * dt/dx but I will just have to study a bit more

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