Question

Qual o resultado da derivada da função f(x) e².cos 3x ?

Obs.: e²x ("e" elevado a 2x).

Answer 1

$Utilizando~a~Regra~do~Produto\\\\\\\frac{df(x)}{dx}~=~\left(e^{2x}\right)'~.~cos(3x)~+~e^{2x}~.~(~cos(3x)~)'\\\\\\Seja:\\u(x)~=~2x\\w(x)~=~3x\\\\\\Utilizando~a~Regra~da~Cadeia:\\\\\frac{df(x)}{dx}~=~\left(e^{u(x)}\right)'~.~cos(w(x))~+~e^{u(x)}~.~(~cos(w(x)~)'\\\\\\\frac{df(x)}{dx}~=~\left(~(e^{u(x)})'.(u(x))'~\right).cos(w(x))~+~e^{u(x)}.\left(~cos(w(x))'.(w(x))'~\right)\\\\\\\frac{df(x)}{dx}~=~\left(~(e^{u(x)})'.(u(x))'~\right).cos(w(x))~+~e^{u(x)}.\left(~cos(w(x))'.(w(x))'~\right)$

$\frac{df(x)}{dx}~=~\left(~e^{u(x)}.(2)~\right).cos(w(x))~+~e^{u(x)}.\left(~-sen(w(x)).(3)~\right)\\\\\\\frac{df(x)}{dx}~=~2.e^{u(x)}.cos(w(x))~-~3.e^{u(x)}.sen(w(x))\\\\\\\frac{df(x)}{dx}~=~2.e^{2x}.cos(3x)~-~3.e^{2x}.sen(3x)\\\\\\\boxed{\frac{df(x)}{dx}~=~e^{2x}\,.\,\left(~2cos(3x)~-~3sen(3x)~\right)}$