Question

Derivada de
$y = \frac{e^x^2}{x}$

OBS : Esse 2 é uma potência do x

Answer 1

$y=\dfrac{e^{x^{2}}}{x}$

Derivando usando a Regra da Cadeia, combinada com a Regra do Quociente:

$\dfrac{dy}{dx}=\dfrac{d}{dx}\!\left(\dfrac{e^{x^{2}}}{x} \right )\\\\\\ \dfrac{dy}{dx}=\dfrac{\frac{d}{dx}(e^{x^{2}})\cdot x-e^{x^{2}}\cdot \frac{d}{dx}(x)}{x^{2}}\\\\\\ \dfrac{dy}{dx}=\dfrac{e^{x^{2}}\cdot \frac{d}{dx}(x^{2})\cdot x-e^{x^{2}}\cdot 1}{x^{2}}\\\\\\ \dfrac{dy}{dx}=\dfrac{e^{x^{2}}\cdot 2x\cdot x-e^{x^{2}}\cdot 1}{x^{2}}\\\\\\ \dfrac{dy}{dx}=\dfrac{e^{x^{2}}\cdot 2x^{2}-e^{x^{2}}\cdot 1}{x^{2}}\\\\\\ \boxed{\begin{array}{c}\dfrac{dy}{dx}=\dfrac{e^{x^{2}}\cdot (2x^{2}-1)}{x^{2}} \end{array}}$

Answer 2

$\sf \displaystyle y=\frac{e^{x\:2}}{x}\\\\\\\frac{d}{dx}\left(\frac{e^{x\cdot \:2}}{x}\right)\\\\\\{Aplique\:a\:regra\:do\:quociente}:\quad \left(\dfrac{f}{g}\right)=\frac{f\:'\cdot g-g'\cdot f}{g^2}\\\\\\=\frac{\frac{d}{dx}\left(e^{x\cdot \:2}\right)x-\frac{d}{dx}\left(x\right)e^{x\cdot \:2}}{x^2}\\\\\\=\frac{e^{x\cdot \:2}\cdot \:2x-1\cdot \:e^{x\cdot \:2}}{x^2}\\\\\\\to \boxed{\sf =\frac{2e^{2x}x-e^{2x}}{x^2}}$