## jaersyn 3 years ago What is the derivative of: (secu)^(n-2)? Is it (n-2)(secu)^(n-3) tanu?

1. jaersyn

n is an integer

2. hsmt

hi. oh, ok

3. hsmt

4. jaersyn

just a variable

5. hsmt

not a function of x?

6. hsmt

ok, u is not a function of x, then i think you are gonna want to check the exponent of sec u

7. hsmt

still there?

8. jaersyn

yeah

9. jaersyn

i got it

10. hsmt

in case you get this, everything looks ok except the exponent on sec u. the n-2 out front is good, the sec^(n-3) times the sec u tan u (chain rule) is what is affecting the exponent

11. hsmt

do you see what the exponent is now for sec u?

12. jaersyn

it's n-3 no?

13. hsmt

the n-2 in front is correct. you also have product of secu^(n-3) and then there is the chain rule that multiplies by secutanu. so, it is not n-3

14. hsmt

do you see what i am trying to point out? feel free to ask any questions

15. hsmt

are you still having difficulty?

16. jaersyn

oh pellet

17. jaersyn

you are correct

18. jaersyn

it would still be sec^(n-2)

19. hsmt

niiice

20. jaersyn

(n-2)(secu)^(n-3) times secutanu which equals (n-2)(secu)^(n-2) times tanu

21. hsmt

very cool you got it

22. jaersyn

now i have to rewrite this wholeeeeeeeeeee thing......... thanks

23. hsmt

dontcha just hate the whole rewriting thing? If there is nothing else i can help with, i will bid you adieu

24. hsmt

take care