′(x)=(5⋅x7−x6+xcos(x))′==(5⋅x7−x6+x)′⋅cos(x)−(5⋅x7−x6+x)⋅(cos(x))′cos2(x)==((5⋅x7−x6)′+1)⋅cos(x)−(5⋅x7−x6+x)⋅(−sin(x))cos2(x)==((5⋅x7)′−(x6)′+1)⋅cos(x)−(5⋅x7−x6+x)⋅(−sin(x))cos2(x)==(5⋅(x7)′−6⋅x5+1)⋅cos(x)−(5⋅x7−x6+x)⋅(−sin(x))cos2(x)==(35⋅x6−6⋅x5+1)⋅cos(x)−(5⋅x7−x6+x)⋅(−sin(x))cos2(x)
Ответ:
f′(x)=(35⋅x6−6⋅x5+1)⋅cos(x)−(5⋅x7−x6+x)⋅(−sin(x))cos2(x)