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Finding derivative of sin^-1 (x^2 + 1)

  1. Apr 17, 2010 #1
    1. The problem statement, all variables and given/known data

    Diffentiate [tex]f(x)=\sin^{-1}(x^2+1)[/tex]

    2. Relevant equations



    3. The attempt at a solution

    [tex]f'(x)=(-1)(2x)\sin^{-2} (x^2+1)\cos (x^2+1)[/tex]

    am i correct ? But wolfram alpha is giving me something else

    http://www.wolframalpha.com/input/?i=differentiate+sin^(-1)+(x^2+1)
     
  2. jcsd
  3. Apr 17, 2010 #2

    tiny-tim

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    Hi thereddevils! :smile:

    (try using the X2 tag just above the Reply box :wink:)

    You're misreading the (admittedly slightly misleading :rolleyes:) special inverse trig notation …

    you've correctly differentiated f(x) = 1/sin(x2+1) :smile:,

    but the question means f(x) = arcsin(x2+1) :wink:

    (see eg http://mathworld.wolfram.com/InverseTrigonometricFunctions.html" [Broken] doesn't help at all :frown:)
     
    Last edited by a moderator: May 4, 2017
  4. Apr 17, 2010 #3
    Re: differentiation

    thanks tiny , but whats that tag for ? seems that it makes my font smaller .
     
  5. Apr 17, 2010 #4

    tiny-tim

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    Should be smaller, but higher up: 222 :rolleyes: … wheee! :tongue2:

    Are you using the same tag as me? …

    it's on the second row (the one that starts B I U …), six from the end :smile:
     
  6. Apr 17, 2010 #5
    Re: differentiation

    x2 , wow never know it can be done that way , interesting !!

    Vo

    Why is there no latex tags in this forum ? I will need to type the tags myself which is sometimes troublesome :biggrin:
     
  7. Apr 17, 2010 #6

    tiny-tim

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    The ∑ tag (at the end of the line) gives you lots of latex symbols, and the first one you click also gives you [noparse][tex] and [/tex][/noparse] :wink:
     
  8. Apr 19, 2010 #7
    Re: differentiation

    Table of Derivatives:
    [PLAIN]https://dl.dropbox.com/u/4645835/MATH/derv_arcsin.gif [Broken]
     
    Last edited by a moderator: May 4, 2017
  9. Apr 19, 2010 #8
    Re: differentiation

    Remember the Chain rule, reddevils?

    which says let

    y = y(u(x)) where derivative is

    [tex]\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}[/tex]

    which you properly also know as

    [tex](f \circ g)'(x) = f'(g(x)) \cdot g'(x)[/tex]

    since you know that

    [tex] f(u) = sin^{-1}(u)[/tex]

    and

    [tex]u = x^2+1[/tex]

    then its up to you to use formula above correctly :D

    Sincerely
    Susanne
     
  10. Apr 20, 2010 #9
    Re: differentiation

    thanks Susan .
     
  11. Apr 21, 2010 #10
    Re: differentiation

    You are welcome!
     
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