Question

y= 7sin3x+2cos5x, x0=п/3 ; y' (x0)-?

Answer 1

$y'_x=7*3*cos(3x)-2*5sin(5x)$

 

$y'_x(x_0)=7*3*cosleft(3*frac{pi}{3}right)-2*5sinleft(5*frac{pi}{3}right)$

 

$3*7*cosleft(3*frac{pi}{3}right)-2*5sinleft(5*frac{pi}{3}right)=3*7*cospi-10*sinleft(pi+frac{2pi}{3}right))$

 

$3*7*cospi-10*sinleft(pi+frac{2pi}{3}right)=$

 

$-21-10*(sinpicosfrac{2pi}{3}+cospisinfrac{2pi}{3})=-21-10*(-sinfrac{2pi}{3})$

 

$-21+10*frac{sqrt{3}}{2}=-21+5sqrt{3}$