Question

f(x)= 2sen2(x)+2cos2(x)+x

Answer 1

Resposta:

essa é a mesma conta que fiz na mesma questão

f'(x)=4ens+4cos+1

Explicação passo-a-passo:

f(x)=2sen2(x)+2cos2(x)+x

f(x)=2sen×2x+2cos×2x+x

f'(x)=d/dx (2sen×2x+2cos ×2x+x)

2sen×2x

4senx

f'(x)=d/dx (2sen×2x+2cos×2x+x)

f'(x)=d/dx(4senx+2cos×2x+x)

f'(x)=d/dx(2sen×2x+2cos×2x+x)

4cos×2x

4cosx

f'(x)=d/dx(2sen×2x+2cos×2x+x)

f'(x)=d/dx(4senx+4cosx+x)

f'(x)=d/dx(4senx)+d/dx(4cosx)+d/dx(x)

d/dx(4senx)

4sen

f'(x)=d/dx(4senx)+d/dx(4cosx)+d/dx(x)

f'(x)=4sen+d/dx(4cosx)+d/dx(x)

f'(x)=d/dx(4senx)+d/dx(4cosx)+d/dx(x)

d/dx(4cosx)

4cos

f'(x)=d/dx(4senx)+d/dx(4cosx)+d/dx(x)

f'(x)=4sen+4cos+d/dx(x)

f'(x)=d/dx(4senx)+d/dx(4cosx)+d/dx(x)

1

f'(x)=d/dx(4senx)+d/dx(4cosx)+d/dx(x)

f'(x)=4sen+4cos+1

f'(x)=4ens+4cos+1

Resposta

f'(x)=4ens+4cos+1