Question

Ajuda...

Como derivo essa função. ?
1 / (x2-3x-2) ^5

Answer 1

$y=\dfrac{1}{(x^{2}-3x-2)^{5}}$


Regra da Cadeia: Fazendo,

$\bullet\;\;$ [/tex]y=\dfrac{1}{u};[/tex]

onde $u=(x^{2}-3x-2)^{5};$


$\bullet\;\;$ $u=v^{5};$

onde$v=x^{2}-3x-2$


temos que

$\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dv}\cdot \dfrac{dv}{dx}\\ \\ \\ \dfrac{dy}{dx}=\dfrac{d}{du}\left(\dfrac{1}{u} \right )\cdot \dfrac{d}{dv}(v^{5})\cdot \dfrac{d}{dx}(x^{2}-3x-2)\\ \\ \\ \dfrac{dy}{dx}=\dfrac{d}{du}(u^{-1})\cdot \dfrac{d}{dv}(v^{5})\cdot \dfrac{d}{dx}(x^{2}-3x-2)\\ \\ \\ \dfrac{dy}{dx}=(-1)\cdot u^{-1-1}\cdot 5v^{5-1}\cdot (2x-3)\\ \\ \\ \dfrac{dy}{dx}=-u^{-2}\cdot 5v^{4}\cdot (2x-3)$


Voltando à variável $x,$ temos

$\dfrac{dy}{dx}=-(v^{5})^{-2}\cdot 5v^{4}\cdot (2x-3)\\ \\ \\ \dfrac{dy}{dx}=-v^{-10}\cdot 5v^{4}\cdot (2x-3)\\ \\ \\ \dfrac{dy}{dx}=-5v^{-10+4}\cdot (2x-3)\\ \\ \\ \dfrac{dy}{dx}=-5v^{-6}\cdot (2x-3)\\ \\ \\ \dfrac{dy}{dx}=-5\,(x^{2}-3x-2)^{-6}\cdot (2x-3)\\ \\ \\ \boxed{\dfrac{dy}{dx}=-\dfrac{5\,(2x-3)}{(x^{2}-3x-2)^{6}}}$