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Differentiate Inverse Cos – Proof

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How to differentiate cos-1x

y=cos-1x
Bring the cos across
cosy = x
Differentiate both sides, remember when differentiating y time by dy/dx
-sin(y) dy/dx = 1
dy/dx = -1/siny

However we want to get the differential in terms of x, to do this we can use the identity
sin2t+cos2t = 1
so
sint = √(1 – cos2t)

putting this into our expression for dy/dx we get

dy/dx = -1/√(1-cos2y)
but cosy = x so

dy/dx =- 1/√(1-x2)

  1. Denaya Lesa
    January 14th, 2009 at 09:20 | #1

    He he he, I think you must take the following conclusions:

    the differentiate of cos-1(x) is dy/dx = – 1/√(1-x2)

    while

    the differentiate of sin-1(x) is dy/dx = 1/√(1-x2)

    How about you all?

  2. Denaya Lesa
    January 14th, 2009 at 09:42 | #2

    Hmm, if you want to know an application of both cos-1(x) and sin-1(x) simultaneously, please read rohedi’s comment on

    http://unapologetic.wordpress.com/2008/10/10/the-exponential-differential-equation/.

  3. Nadya Fermega
    January 14th, 2009 at 09:52 | #3

    @Denaya

    No no no, you are wrong Denaya, rohedi’s comment corresponds to your purpose is at this address

    http://unapologetic.wordpress.com/2008/10/13/sine-and-cosine/

    when Mr. Rohedi solves f”+f=0

    I think you have done big blunder, huahuahau…

  4. Nadya Fermega
    January 14th, 2009 at 09:53 | #4

    @Denaya

    No no no, you are wrong Denaya, rohedi’s comment corresponds to your purpose is at this address

    http://unapologetic.wordpress.com/2008/10/13/sine-and-cosine/

    when Mr. Rohedi solves f”+f=0

    I think you have done big blunder, huahuahau…sorry honey.

  5. Barbara
    February 8th, 2009 at 22:07 | #5

    You’ve missed the minus!!!
    d/dy(arccos) = – 1/ sq(1-x^2)

  6. February 9th, 2009 at 17:51 | #6

    oops sorry thanks for pointing it out
    dave

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