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anonymous

  • one year ago

f(x) = 3 + x^2 + tan (pix/2)

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  1. anonymous
    • one year ago
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    Any one can help me with this derivative of an inverse

  2. sweetburger
    • one year ago
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    wait so your looking to take the derivative of this equation or what exactly?

  3. anonymous
    • one year ago
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    yes Im in the last part but I dont get quite the exact answer

  4. anonymous
    • one year ago
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    I get (pi sec^2 pix/2)/(2) + 2x

  5. freckles
    • one year ago
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    why do you say derivative of inverse ?

  6. freckles
    • one year ago
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    Do you mean just find derivative?

  7. sweetburger
    • one year ago
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    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
    • one year ago
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    so it would look like 2x+sec^2x(pix/2)(pi/2) I think

  9. anonymous
    • one year ago
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    yes, i evaluated at 0 but i get pi^2/4

  10. sweetburger
    • one year ago
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    x+sec^2(pix/2)(pi/2) is the correct value I accidentally left in x above.

  11. freckles
    • one year ago
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    I'm still confused why you said inverse

  12. freckles
    • one year ago
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    are you wanting to find \[(f^{-1})'(a)\] for some number a

  13. anonymous
    • one year ago
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    yes freckles

  14. freckles
    • one year ago
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    what is the a

  15. anonymous
    • one year ago
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    I used a couple of engines and I get (pi*sec^2(pix/2)(2))/ + 2x

  16. anonymous
    • one year ago
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    a = 3

  17. sweetburger
    • one year ago
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    @Eco1 you you are right I made a typo and left out the 2 in front of the x

  18. anonymous
    • one year ago
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    So I evaluate what I wrote above at zero but I dont get pi/2, which is the right asnwer

  19. freckles
    • one year ago
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    \[(f^{-1})'(3)=\frac{1}{f'(f^{-1}(3))} \\ f^{-1}(3)=? \implies f(?)=3 \\ \text{ so we need \to find ? such that }3=3+?^2+\tan(\frac{\pi ?}{2}) \\ \text{ and yes } f'(x)=0+2x+\frac{\pi}{2} \sec^2(\frac{\pi}{2}x)\]

  20. freckles
    • one year ago
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    that should be pretty easy to find ?

  21. freckles
    • one year ago
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    because we know both tan(0)=0 and 0^2=0

  22. freckles
    • one year ago
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    so ?=0

  23. freckles
    • one year ago
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    \[(f^{-1})'(3)=\frac{1}{f'(0)}\]

  24. freckles
    • one year ago
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    last thing for you to do is just to plug in 0 into your f' you found

  25. anonymous
    • one year ago
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    I did multiple times but I dont get the right answer

  26. freckles
    • one year ago
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    \[f'(0)=2(0)+\frac{\pi}{2}[\sec(\frac{\pi}{2}(0))]^2\]

  27. freckles
    • one year ago
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    well first the inside of that sec( ) thing you can be simplified pi/2*0=0 so sec(0)=?

  28. freckles
    • one year ago
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    \[(f^{-1})'(3)=\frac{1}{f'(0)}=\frac{1}{2(0)+\frac{\pi}{2}[\sec(0)]^2}\]

  29. freckles
    • one year ago
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    you know cos(0)=1 so sec(0)=?

  30. freckles
    • one year ago
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    hint sec and cos are reciprocal functions of each other

  31. anonymous
    • one year ago
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    reciprocal is synonym of inverse?

  32. freckles
    • one year ago
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    it depends what inverse you are talking about multiplicative inverse yeah

  33. freckles
    • one year ago
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    like sec(x)=1/cos(x) so sec(0)=1/cos(0)

  34. freckles
    • one year ago
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    1/cos(x) is reciprocal of cos(x)

  35. freckles
    • one year ago
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    some people just like to call it the flipping of

  36. anonymous
    • one year ago
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    but it is sec(0)^2 not sec(0)

  37. anonymous
    • one year ago
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    I meant sec^2(0)

  38. freckles
    • one year ago
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    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
    • one year ago
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    1/(a/b) means just flip the number in the denominator

  40. freckles
    • one year ago
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    so 1/(a/b) is b/a

  41. freckles
    • one year ago
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    are you cool?

  42. freckles
    • one year ago
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    still processing?

  43. anonymous
    • one year ago
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    Thanks man, or girl?

  44. anonymous
    • one year ago
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    I just made an algebra error

  45. anonymous
    • one year ago
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    I got it now, but this problem is a pain in the behind

  46. freckles
    • one year ago
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    I might be a she. And also it gets way more fun with more practice.

  47. anonymous
    • one year ago
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    I think my algebra is rusty

  48. freckles
    • one year ago
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    it seems like everyone is working out their algebra kinks in calculus

  49. freckles
    • one year ago
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    so no worries

  50. anonymous
    • one year ago
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    tell me about it

  51. anonymous
    • one year ago
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    Any one out there who can give me a hand doing integrals

  52. freckles
    • one year ago
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    post a new question so people can see it :)

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