Asked by Ryan
Derivative of (4 secx tanx)
Answers
Answered by
mathhelper
y = 4secx tanx
by the product rule ....
dy/dx = 4secx(sec^2 x) + 4tanx(secxtanx)
= 4sec^3 x + 4secx tan^2x
= 4secx(sec^2 x + tan^2 x) , but sec^2 x - 1 = tan^2 x
= 4secx ( sec^2 x + sec^2x + 1)
= 4secx(2sec^2 x + 1) or 8sec^3 x + 4secx
by the product rule ....
dy/dx = 4secx(sec^2 x) + 4tanx(secxtanx)
= 4sec^3 x + 4secx tan^2x
= 4secx(sec^2 x + tan^2 x) , but sec^2 x - 1 = tan^2 x
= 4secx ( sec^2 x + sec^2x + 1)
= 4secx(2sec^2 x + 1) or 8sec^3 x + 4secx
Answered by
Anonymous
d/dx [ 4 * 1/cos x * sin x /cos x ]
= 4 d/dx [ sin x/cos^2 x}
=4 [ cos^3 x - sin x(2 cos x * -sin x) ] / cos^4 x
=4 [ cos^3 x - sin 2x + sin^2 x ] / cos^4 x
= 4 d/dx [ sin x/cos^2 x}
=4 [ cos^3 x - sin x(2 cos x * -sin x) ] / cos^4 x
=4 [ cos^3 x - sin 2x + sin^2 x ] / cos^4 x
Answered by
oobleck
4 [ cos^3 x - sin x(2 cos x * -sin x) ] / cos^4 x
= (4cos^2x + 2sin^2x)/cos^3x
= (2cos^2x+2)/cos^3x
= (4cos^2x + 2sin^2x)/cos^3x
= (2cos^2x+2)/cos^3x
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