## baldymcgee6 3 years ago Derivatives of parametric curves: http://screencast.com/t/x5c5hBqr

1. baldymcgee6

@hartnn, @klimenkov

2. baldymcgee6

|dw:1351711571629:dw|

3. hartnn

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

4. klimenkov

By what parameter you want to differentiate cot?

5. hartnn

i think x, isn't it ?

6. baldymcgee6

|dw:1351712169898:dw|

7. hartnn

u got what i did ?

8. baldymcgee6

when t = pi/4

9. baldymcgee6

I'm still confused how you got d/dt(-cot t) * dt/dx

10. hartnn

|dw:1351712487011:dw|

11. hartnn

$$\huge \frac{d}{dx}f(t)=\frac{d}{dt}f(t) \frac{dt}{dx}$$

12. baldymcgee6

oohhh.. okay... so we can write:|dw:1351712793765:dw|

13. baldymcgee6

|dw:1351712929460:dw|

14. hartnn

seems correct, now u can put t=pi/4

15. baldymcgee6

|dw:1351713083407:dw|

16. hartnn

yup.

17. baldymcgee6

hmmmm... okay thanks so much for the help... you're the best!

18. hartnn

not really... welcome ^_^

19. baldymcgee6

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