F(r)= 1- Sin(2r) + 2sec2 (r)so Far I Have -cos(2r)

mounampesiyadhe

From: US
Posts: 4
Votes: 12
F(r)= 1- sin(2r) + 2sec2 (r) so far I have -cos(2r) + 4tan (r) is this right and whatdo 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?

 

Answer No.1

FamousBob

From: US
Posts: 14
Votes: 28
( a ) f( r ) = 1 - sin ( 2r ) + 2sec2 ( 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

Answer No.2

amat009ify

From: US
Posts: 33
Votes: 99
F(r) = 1 - sin2r + 2(secr)^2 f ' (r) = - 2cos2r +4(secr)(secrtanr) 2) k(x) = [7x^2 - 9] / [x^3+1] use thequotient 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

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