Question

Calcule y´´ (derivada sucessiva)

$y=x^3.e^x+cos(3x)$

Answer 1

$\\ y = x^3e^x+cos(3x) \\ \\ y' = x^3'e^x+x^3e^x'+[cos(3x)]'*(3x)' \\ \\ y' = 3x^2e^x +x^3e^x -Sen(3x)*3$

---------------------------------------


$\\ y' = 3x^2e^x +x^3e^x -3Sen(3x) \\ \\ y'' = (3x^2)'e^x+(3x^2)e^x'+(x^3)'e^x+(x^3)*e^x'+[-3Sen(3x)]'*(3x)' \\ \\ y'' = 6xe^x+3x^2e^x+3x^2e^x+x^3e^x-3Cos(3x)*3 \\ \\ y'' = 6xe^x+23x^2e^x+x^3e^x-9*Cos(3x)$

Answer 2

$\sf \displaystyle \:y=x^3\cdot \:e^x+cos\left(3x\right)\\\\\\\frac{d}{dx}\left(x^3e^x+cos \left(3x\right)\right)\\\\\\=\frac{d}{dx}\left(x^3e^x\right)+\frac{d}{dx}\left(cos \left(3x\right)\right)\\\\\\{Aplicar\:a\:regra\:da\:soma/diferença}\to :\quad \left(f\pm g\right)'=f\:'\pm g'\\\\\\\to \boxed{\sf =3x^2e^x+e^xx^3-sin \left(3x\right)\cdot \:3 }$