anonymous one year ago f(x) = 3 + x^2 + tan (pix/2)

1. anonymous

Any one can help me with this derivative of an inverse

2. sweetburger

wait so your looking to take the derivative of this equation or what exactly?

3. anonymous

yes Im in the last part but I dont get quite the exact answer

4. anonymous

I get (pi sec^2 pix/2)/(2) + 2x

5. freckles

why do you say derivative of inverse ?

6. freckles

Do you mean just find derivative?

7. sweetburger

so 3 gets deleted x^2 turns into 2x and tan(pix/2) = sec^2x(pix/2) times the derivative of pix/2 which i believe equates to pi/2. I am not sure if this is totally correct so ya check with someone else.

8. sweetburger

so it would look like 2x+sec^2x(pix/2)(pi/2) I think

9. anonymous

yes, i evaluated at 0 but i get pi^2/4

10. sweetburger

x+sec^2(pix/2)(pi/2) is the correct value I accidentally left in x above.

11. freckles

I'm still confused why you said inverse

12. freckles

are you wanting to find $(f^{-1})'(a)$ for some number a

13. anonymous

yes freckles

14. freckles

what is the a

15. anonymous

I used a couple of engines and I get (pi*sec^2(pix/2)(2))/ + 2x

16. anonymous

a = 3

17. sweetburger

@Eco1 you you are right I made a typo and left out the 2 in front of the x

18. anonymous

So I evaluate what I wrote above at zero but I dont get pi/2, which is the right asnwer

19. freckles

$(f^{-1})'(3)=\frac{1}{f'(f^{-1}(3))} \\ f^{-1}(3)=? \implies f(?)=3 \\ \text{ so we need \to find ? such that }3=3+?^2+\tan(\frac{\pi ?}{2}) \\ \text{ and yes } f'(x)=0+2x+\frac{\pi}{2} \sec^2(\frac{\pi}{2}x)$

20. freckles

that should be pretty easy to find ?

21. freckles

because we know both tan(0)=0 and 0^2=0

22. freckles

so ?=0

23. freckles

$(f^{-1})'(3)=\frac{1}{f'(0)}$

24. freckles

last thing for you to do is just to plug in 0 into your f' you found

25. anonymous

I did multiple times but I dont get the right answer

26. freckles

$f'(0)=2(0)+\frac{\pi}{2}[\sec(\frac{\pi}{2}(0))]^2$

27. freckles

well first the inside of that sec( ) thing you can be simplified pi/2*0=0 so sec(0)=?

28. freckles

$(f^{-1})'(3)=\frac{1}{f'(0)}=\frac{1}{2(0)+\frac{\pi}{2}[\sec(0)]^2}$

29. freckles

you know cos(0)=1 so sec(0)=?

30. freckles

hint sec and cos are reciprocal functions of each other

31. anonymous

reciprocal is synonym of inverse?

32. freckles

it depends what inverse you are talking about multiplicative inverse yeah

33. freckles

like sec(x)=1/cos(x) so sec(0)=1/cos(0)

34. freckles

1/cos(x) is reciprocal of cos(x)

35. freckles

some people just like to call it the flipping of

36. anonymous

but it is sec(0)^2 not sec(0)

37. anonymous

I meant sec^2(0)

38. freckles

if cos(0)=1 then sec(0)=1/cos(0)=1/1=1 if you square both sides of cos(0)=1 you get still cos^2(0)=1 so sec^2(0)=1 $(f^{-1})'(3)=\frac{1}{f'(0)}=\frac{1}{0+\frac{\pi}{2}[1]^2}=\frac{1}{\frac{\pi}{2}}$

39. freckles

1/(a/b) means just flip the number in the denominator

40. freckles

so 1/(a/b) is b/a

41. freckles

are you cool?

42. freckles

still processing?

43. anonymous

Thanks man, or girl?

44. anonymous

I just made an algebra error

45. anonymous

I got it now, but this problem is a pain in the behind

46. freckles

I might be a she. And also it gets way more fun with more practice.

47. anonymous

I think my algebra is rusty

48. freckles

it seems like everyone is working out their algebra kinks in calculus

49. freckles

so no worries

50. anonymous

51. anonymous

Any one out there who can give me a hand doing integrals

52. freckles

post a new question so people can see it :)