Question

Нужно найти производную срочно
1)F(x)=5sin x cos x
2)f(x)=cos(4-3x)
3)f(x)=ctg(2-5x)

Answer 1

1. y′= (5·sin(x))(cos(x))′=
=5·(sin(x))(cos(x))′=
=5·(x))(cos(x)′·cos(x))(cos(x)=5·0·cos(x))(cos(x)=
=0
2. y′= (cos(4−(3·x)))′=(4−(3·x))′·(−1)·sin(4−(3·x))=
=−(4−(3·x))′·sin(4−(3·x))=
=−((4)′−(3·x)′)·sin(4−(3·x))=−(0−(3·(x)′))·sin(4−(3·x))=
=1·sin(4−(3·x))·3·(x)′=1·sin(4−(3·x))·3·1=
=3·sin(4−(3·x))
3. y′= (ctg(2−(5·x)))′=
=−
(2−(5·x))′
sin2(2−(5·x))2
=
=−
(2)′−(5·x)′
sin2(2−(5·x))2
=−
0−(5·(x)′)
sin2(2−(5·x))2
=
=−
(−1)·5·(x)′
sin2(2−(5·x))2
=−
(−1)·5·1
sin2(2−(5·x))2
=
=
5
sin2(2−(5·x))2


проверяйте наывсякий случай

Answer 2

????....

Answer 3

1)F(x)=5sinx cosx
 (uv)´=u´v+uv´
   F´(x)=5(cosxcosx+sinx(-sinx)=5(cos²x-sin²x)=5cos2x
((sinx)´=cosx, (cosx)´=-sinx, cos²x-sin²x=cos2x)
2)f(x)=cos(4-3x)
   f´(x)=-sin(4-3x) . (4-3x)´=-sin(4-3x) . (-3)=3sin(4-3x)
(( F(g(x))´=F´(g(x)).g´(x))
3)f(x)=ctg(2-5x)
   f´(x)=(-5).(-1/(sin²x))=5/sin²x, x≠kπ,k∈Z
  sin(2-5x)≠0, 2-5x≠kπ, 5x≠2-kπ, x≠(2-kπ)/5