## anonymous 4 years ago inverse and inverse trignometric functions find d/dx(arctan^2(sqrt(x))

1. anonymous

hi krypton.. you still there? I think i might be able to help if you are

2. anonymous

yea dear

3. anonymous

consider the derivative of arctan^2(u) where u=sqrt x and use the chain rule...

4. anonymous

sorry. let's start with the derivative of (tan-1u)^2. what is the derivative of this little beastie. lol there we go...

5. anonymous

u mean?

6. anonymous

what is the derivative of (f(x))^2?

7. anonymous

it would be something like 2(f(x))times f prime of x

8. anonymous

in this case, f(u)=tan-1u, where u=sqrt of x. does this make sense?

9. anonymous

questions are good here. the more you ask, the more i can help... :)

10. anonymous

yea

11. anonymous

so what is the derivative of tan-1u?

12. anonymous

you can give me your best guess if you like

13. anonymous

hmm actually i wasnt in the class when the topic was taught, but i thinkis =1/srt 1-x^2?

14. anonymous

sorry 1/1+x^2

15. anonymous

that is totally correct. :) now replace the x with u so that we can use the chain rule where u=1/sqrtx.

16. anonymous

17. anonymous

no, we have a power of tan-1. so the answer is 2(tan-1(sqrtx))(1/(1+x))1/2sqrtx

18. anonymous

is that making sense to you?

19. anonymous

still there?

20. anonymous

ok

21. anonymous

so what will be my final answer?

22. anonymous

it is above

23. anonymous

can u help me with one other questionÉ

24. anonymous

if i can, i would be happy to

25. anonymous

find $find d/dx (\sqrt{1-x ^{2}}+1/2\cos ^{-1}(x) and verify whether or \not this is equal \to (2x+1)d/dx \sin ^{-1}(-x)$

26. anonymous

ok. any ideas where to start?

27. anonymous

oh... and i didnt get anything beyond the arrow... sorry... but i think i have enough to figure it out. :)

28. anonymous

hmm i dont have an idea

29. anonymous

ok, well, it is a rather large function, but it is a quotient (hint). So where might we start with this lovely little thing?

30. anonymous

hmm..no idea lol

31. anonymous

dont know anytin about the topic yet

32. anonymous

it is ok... remember the quotient rule d(f/g)/dx=[(fprime)g-(gprime)f]/g^2?

33. anonymous

in this case the numerator is f=sqrt(1-x^2) +1 and g=2cos-1(x)

34. anonymous

do you know the product rule?

35. anonymous

yes

36. anonymous

phew...!! i am so glad. i thought for a sec we were outta options. lol

37. anonymous

ok

38. anonymous

look at f(x)=sqrt(1-x^2)+1 and g(x)=(2cos-1(x))^-1. does this make sense?

39. anonymous

hahaha i can still remember my calculus 1, this calculus 2

40. anonymous

yes it does

41. anonymous

glad to hear it... calc I is totally needed here... good, it makes sense. now we need fprime and gprime and i think you are pretty much homefree. so what is fprime and what is gprime?

42. anonymous

just algebra after you get the derivatives.

43. anonymous

so i should get the derivative of each of them separatelyÉ