Question

derivada de y=cos(x^2+2x-1)-3senx

Answer 1

d/dx (y(x))= d/dx (cos(x^2+2x-1)-3sen(x))

d/dx(y(x))= d/dx (cos(1-2x-x^2)-3sen(x))

logo a derivada de y é y':

y'(x)= d/dx (cos(1-2x-x^2)) - 3 d/dx (sen(x))

Aplicando a regra da cadeia temos:

d/dx (cos(-x^2-2x+1)) = ( dcos(u) / du ) * (du/dx) ;
onde:
u= -x^2-2x+1 e d/du(cos(u))= - sen(u)

Assim,

y'(x) = -3 (d/dx (sen(x))) + (-d/dx(1-2x-x^2)sen(1-2x-x^2))

y'(x) = -3 (d/dx (sen(x))) - d/dx(1) - 2d/dx(x) - d/dx(x^2) sen(1-2x-x^2)

y'(x) = -(d/dx(1) - 2(d/dx(x)) - d/dx(x^2))sen(1-2x-x^2)-3cosx

y'(x) = -3cosx * (-2d/dx(x) - d/dx(x^2))* sen(1-2x-x^2)

y'(x)= 2(x+1)sen(-x^2-2x+1)-3cos(x)

ou

y'(x) = -3cosx - sen(1-2x-x^2)*(-2-2x)
fim

Answer 2

y = cos(x² + 2x - 1) - 3senx

y' = -sen(x² + 2x - 1) . (x² - 2x - 1)' - 3cosx . x'

y' = -sen(x² - 2x - 1) (2x - 2) - 3cosx . 1

y' = -2(x - 1)sen(x² - 2x - 1) - 3cosx