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fedёk1324
@fedёk1324
August 2022
2
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((1-cos4x)/sin4x)' - производная
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Answers & Comments
mikael2
Y=((1-cos4x)/sin4x) y'=(u/v)'=1/v²[u'v-v'u]
u=1-cos4x u'=4sin4x
v=sin4x v'=4cos4x
((1-cos4x)/sin4x)'=1/sin²4x[4sin4x*sin4x-4cos4x*(1-cos4x)] =
=4/sin²4x[sin²4x+cos²4x-cos4x]= 4(1-cos4x)/sin²4x
1 votes
Thanks 2
sedinalana
Y`=[(1-cos4x)`*sin4x-(sin4x)`(1-cos4x)]/sin²4x=
=(4sin4x*sin4x-4cos4x+4cos²4x)/sin²4x=[4(sin²4x+cos²4x)-4cos4x]/sin²4x=
=(4-4cos4x)/sin²4x=4(1-cos4x)/sin²4x=2sin²2x/(4sin²2xcos²2x)=
=1/(2cos²2x)
1 votes
Thanks 1
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Answers & Comments
u=1-cos4x u'=4sin4x
v=sin4x v'=4cos4x
((1-cos4x)/sin4x)'=1/sin²4x[4sin4x*sin4x-4cos4x*(1-cos4x)] =
=4/sin²4x[sin²4x+cos²4x-cos4x]= 4(1-cos4x)/sin²4x
=(4sin4x*sin4x-4cos4x+4cos²4x)/sin²4x=[4(sin²4x+cos²4x)-4cos4x]/sin²4x=
=(4-4cos4x)/sin²4x=4(1-cos4x)/sin²4x=2sin²2x/(4sin²2xcos²2x)=
=1/(2cos²2x)