Ensino Superior ⇒ Derivada Tópico resolvido

[Deleted User 19359]

Determine a derivada f(x) = [tex3]\frac{\\{e^{-x}\arctan e^{x}}}{\tan x}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\frac{\\{e^{-x}\arctan e^{x}}}{\tan x}[/tex3]

[alevini98]

Nesse caso será mais fácil se optarmos pela derivação logarítmica.

[tex3]y=\frac{e^{-x}\arctg e^x}{\tg x}\\\ln y=\ln\left(\frac{e^{-x}\arctg e^x}{\arctg x}\right)\\\ln y=-x+\ln(\arctg e^x)-\ln(\arctg x)\\\frac{1}{y}\frac{dy}{dx}=-1+\frac{1}{\arctg e^x}\cdot\frac{1}{1+e^{2x}}\cdot e^x-\frac{1}{\arctg x}\cdot\frac{1}{1+x^2}\\\frac{dy}{dx}=y\left(-1+\frac{e^x}{(\arctg e^x)(1+e^{2x})}-\frac{1}{(\arctg x)(1+x^2)}\right)\\\frac{dy}{dx}=\frac{e^{-x}\arctg e^x}{\tg x}\left(-1+\frac{e^x}{(\arctg e^x)(1+e^{2x})}-\frac{1}{(\arctg x)(1+x^2)}\right)[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]y=\frac{e^{-x}\arctg e^x}{\tg x}\\\ln y=\ln\left(\frac{e^{-x}\arctg e^x}{\arctg x}\right)\\\ln y=-x+\ln(\arctg e^x)-\ln(\arctg x)\\\frac{1}{y}\frac{dy}{dx}=-1+\frac{1}{\arctg e^x}\cdot\frac{1}{1+e^{2x}}\cdot e^x-\frac{1}{\arctg x}\cdot\frac{1}{1+x^2}\\\frac{dy}{dx}=y\left(-1+\frac{e^x}{(\arctg e^x)(1+e^{2x})}-\frac{1}{(\arctg x)(1+x^2)}\right)\\\frac{dy}{dx}=\frac{e^{-x}\arctg e^x}{\tg x}\left(-1+\frac{e^x}{(\arctg e^x)(1+e^{2x})}-\frac{1}{(\arctg x)(1+x^2)}\right)[/tex3]