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posted by  52010 on 11/5/2009 7:13:34 PM  |  status: Closed  |  Earned Karma: 25

find the derivative

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N/A N/A N/A N/A 11/6/2009 at 1:00:00 PM
Question Details:
f(r)= 1- sin(2r) + 2sec2 (r)
so far I have -cos(2r) + 4tan (r) is this right and what do I do next?
also, for this problem: k(x)= 7x2-9/x3 +1 I have to do the same thing for this one. So far I have (x3+1)(14x) -(7x2-9)(3x) all over (x3+1)2 is this right? What's next?
Tags: Calculus
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Sage
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posted by rimantas on 11/5/2009 7:32:19 PM  |  status: Live
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Response Details:
f(r) = 1 - sin2r + 2(secr)^2
f ' (r) = - 2cos2r  +4(secr)(secrtanr)      
2)  k(x) = [7x^2 - 9] / [x^3+1]   use the quotient rule
    k ' (x) =  [14x(x^3+1) - 3x^2(7x^2-9)] / [x^3+1]
              =  [ 14x^4 + 14x - 21x^4 + 27x^2 ] / [x^3+1]
               =  [ - 7x^4 + 27x^2 + 14x ] / [ x^3+1]
hope this helps
Oracle
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posted by anonymousxyz on 11/5/2009 7:36:22 PM  |  status: Live
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Response Details:


( a )


f( r )  =  1  -  sin ( 2r )  +  2 sec2 ( r )


f '( r )
            =    d / dx [  1  ]    -    d / dx [  sin ( 2r )  ]    +    d / dx [  2 sec2 ( r )  ]

            =    [  0  ]    -    [  cos ( 2r )  ]  *  d / dx [  2r  ]    +    [  4 sec ( r )  ]  *  d / dx  [  sec ( r )  ]

            =    [  0  ]    -     [  cos ( 2r )  ] [  2  ]    +    [  4 sec ( r )  ] [  sec ( r ) tan ( r )  ]

            =    -2 cos ( 2r )    +    4 sec2 ( r ) tan ( r )

( b )


k( x )  =  ( 7x2  -  9 ) / ( x3  +  1 )


u  =  7x2  -  9

v  =  x3  +  1


k'( x )
          =  [  u'v  -  uv'  ]  /  [  v2  ]

          =  [  ( 14x )( x3  +  1 )  -  ( 7x2  -  9 )( 3x2 )  ]  /  [  ( x3  +  1 )2  ]

          =  [  ( 14x4  +  14x )  -  ( 21x4  -  27x2 )  ]  /  [  ( x3  +  1 )2  ]

          =  [  -7x4  +  27x2  +  14x  ]  /  [  ( x3  +  1 )2  ]

          =  ( -7x4  +  27x2  +  14x ) / ( x3  +  1 )2
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