1. f`(x) =( 0.2x⁵ - 3x³ + x + 5)`=0,2·5x⁴-3·3x²+1=x⁴-9x²+1,
2. по формуле производная произведения:
f`(x) = (x²)`·(x-3)+x²·(x-3)`=2x(x-3)+x²=2x²-6x+x²=3x²-6x
или раскроем скобки:
f(x)=x³-3x²
f`(x)=(x³-3x²)`=3x²-3·2x=3x²-6x
3. f`(x) =( -sin x +7cos x - ctg x)`=-cosx-7sinx+(1/sin²x)
4. f`(x) =(√(4x+1) - 4cos2x)`=(4x+1)` ·1/2√(4x+1) -4 (-sin2x)·(2x)`=
= 4/2√(4x+1)+8sin 2x=2/√(4x+1) + 8 sin 2x
задача2
f(x)= 1/2x + sin( x -π/3)
f`(x)=1/2 +cos(x - π/3)
f`(x)=0
1/2 + cos (x - π/3)=0,
cos (x - π/3) =-1/2,
x - π/3=±(arcsin (-1/2) + 2πk, k∈Z
x=π/3 ±(π - π/6) + 2πk, k∈Z
x=π/3 ±(5π/6) + 2πk, k∈Z
Ответ.
x=π/3 ±(5π/6) + 2πk, k∈Z