## krypton Group Title inverse and inverse trignometric functions find d/dx(arctan^2(sqrt(x)) 2 years ago 2 years ago

1. hsmt

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

2. krypton

yea dear

3. hsmt

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

4. hsmt

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

5. krypton

u mean?

6. hsmt

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

7. hsmt

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

8. hsmt

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

9. hsmt

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

10. krypton

yea

11. hsmt

so what is the derivative of tan-1u?

12. hsmt

you can give me your best guess if you like

13. krypton

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

14. krypton

sorry 1/1+x^2

15. hsmt

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

16. krypton

17. hsmt

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

18. hsmt

is that making sense to you?

19. hsmt

still there?

20. krypton

ok

21. krypton

so what will be my final answer?

22. hsmt

it is above

23. krypton

can u help me with one other questionÉ

24. hsmt

if i can, i would be happy to

25. krypton

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. hsmt

ok. any ideas where to start?

27. hsmt

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

28. krypton

hmm i dont have an idea

29. hsmt

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. krypton

hmm..no idea lol

31. krypton

dont know anytin about the topic yet

32. hsmt

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

33. hsmt

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

34. hsmt

do you know the product rule?

35. krypton

yes

36. hsmt

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

37. hsmt

ok

38. hsmt

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

39. krypton

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

40. krypton

yes it does

41. hsmt

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. hsmt

just algebra after you get the derivatives.

43. krypton

so i should get the derivative of each of them separatelyÉ