Question

Calcule a derivada terceira da função f(x) = -cos(x³+3x²-2)

Answer 1

$Aplicando~a~Regra~da~Cadeia\\\\\\Seja:~u(x)~=~x^3+3x^2-2\\\\\\\frac{df(x)}{dx}~=~\frac{df(u)}{du}~.~\frac{u(x)}{dx}\\\\\\\frac{df(x)}{dx}~=~(~-cos(u)~)'~.~(x^3+3x^2-2)'\\\\\\\frac{df(x)}{dx}~=~(~sen(u)~)~.~(3.x^2+3~.~2x^1-0)\\\\\\\boxed{\frac{df(x)}{dx}~=~(3x^2+6x).sen(x^3+3x^2-2)}$

$\frac{df(x)}{dx^2}~=~(~(3x^2+6x).sen(x^3+3x^2-2)~)'\\\\\\Aplicando~a~regra~do~produto:\\\\\\\frac{df(x)}{dx^2}~=~(3x^2+6x)'~.~sen(x^3+3x^2-2)~+~(3x^2+6x)~.~sen(x^3+3x^2-2)'\\\\\\\frac{df(x)}{dx^2}~=~(3~.~2x^1+6)~.~sen(x^3+3x^2-2)~+~(3x^2+6x)~.~sen(x^3+3x^2-2)'\\\\\\Aplicando~a~regra~da~cadeia:\\\\\\Seja:~u(x)~=~x^3+3x^2-2\\\\\\\frac{df(x)}{dx^2}~=~(6x+6)~.~sen(x^3+3x^2-2)~+~(3x^2+6x)~.~sen(u)'.(x^3+3x^2-2)'$

$\frac{df(x)}{dx^2}~=~(6x+6).sen(x^3+3x^2-2)~+~(3x^2+6x)~.~cos(u).(3x^2+6x)\\\\\\\boxed{\frac{df(x)}{dx^2}~=~(6x+6).sen(x^3+3x^2-2)~+~(3x^2+6x)^2~.~cos(x^3+3x^2-2)}$

$\frac{df(x)}{dx^3}~=~(~(6x+6).sen(x^3+3x^2-2)~+~(3x^2+6x)^2~.~cos(x^3+3x^2-2)~)'\\\\\\Aplicando~a~regra~do~produto\\\\\\\frac{df(x)}{dx^3}~=~(6x+6)'.sen(x^3+3x^2-2)~+~sen(x^3+3x^2-2)'.(6x+6)~+~\\\\~~~~~~~~~~~~~(~(3x^2+6x)^2~)'.cos(x^3+3x^2-2)~+~(3x^2+6x)^2.cos(x^3+3x^2-2)'\\\\\\\frac{df(x)}{dx^3}~=~6.sen(x^3+3x^2-2)~+~sen(x^3+3x^2-2)'.(6x+6)~+~\\\\~~~~~~~~~~~~~(~(3x^2+6x)^2~)'.cos(x^3+3x^2-2)~+~(3x^2+6x)^2.cos(x^3+3x^2-2)'\\\\\\Aplicando~a~regra~da~cadeia$

$Seja:~u(x)~=~x^3+3x^2-2~~~~e~~~~w(x)~=~3x^2+6x\\\\\\\frac{df(x)}{dx^3}~=~6.sen(x^3+3x^2-2)~+~sen(u)'.(x^3+3x^2-2)'.(6x+6)~+~\\\\~~~~~~~~~~~~~(~(w)^2~)'.(3x^2+6x)'.cos(x^3+3x^2-2)~+~(3x^2+6x)^2.cos(u)'.(x^3+3x^2-2)'\\\\\\\frac{df(x)}{dx^3}~=~6.sen(x^3+3x^2-2)~+~cos(u).(3x^2+6x).(6x+6)~+~\\\\~~~~~~~~~~~~~2w.(6x+6).cos(x^3+3x^2-2)~+~(3x^2+6x)^2.-sen(u).(3x^2+6x)$

$\frac{df(x)}{dx^3}~=~6.sen(x^3+3x^2-2)~+~\\\\~~~~~~~~~~~~~3.(3x^2+6x).(6x+6).cos(x^3+3x^2-2)~+\\\\~~~~~~~~~~~~~-(3x^2+6x)^3.sen(x^3+3x^2-2)$

A partir daqui você pode tentar simplificar ou colocar em evidencia alguns termos.

Verifique os calculos!