[m]s`=3\cdot cos\frac{t}{a}\cdot (\frac{t}{a})`+0=\frac{3}{a}cos\frac{t}{a}[/m]
2.
y`=-sinx-(1/3)*3cos^2x=cos^2x-sinx
3.
Так как sin2 α =2sin α cos α ⇒ sin α cos α=(1/2)sin2 α
и
[m]sin\frac{x}{8}\cdot cos\frac{x}{8}=\frac{1}{2}sin\frac{x}{4}[/m]
[m]sin\frac{x}{4}\cdot cos\frac{x}{4}=\frac{1}{2}sin\frac{x}{2}[/m]
тогда
[m]f(x)=\frac{1}{4}sin\frac{x}{2}[/m]
[m]f`(x)=\frac{1}{4}cos\frac{x}{2}\cdot (\frac{x}{2})`=\frac{1}{8}cos\frac{x}{2}[/m]
[m]f`(\frac{\pi }{2})=\frac{1}{8}cos\frac{\pi }{4}=\frac{\sqrt{2}}{16}[/m]