Найти значение производной в точке x0 f(x) = (2x)(sin5x), x0=П/2 с подробным решением

У'=2(1+5кос(5х)).если х0=π\2,то:
У'(х0)=2(1+0), у'(х0)=2.

f(x) = 2x * Sin5x
f '(x) = 2(x' * Sin5x + x (Sin5x)' ) = 2(Sin5x + x * Cos5x * (5x)') = 2(Sin5x +
+ 5x*Cos5x)
f '(π/2) = 2(Sin 5π/2 + 5π/2*Cos 5π/2) = 2[Sin(2π+π/2) + 5π/2 * Cos(2π+π/2)]=
= 2(Sin π/2 + 5π/2 * Cos π/2) = 2( 1 + 5π/2 * 0) = 2