Question #88084

Sep 17, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{x} \left(x - 2\right)}{x} ^ 3$

Explanation:

$y = \frac{{e}^{x}}{{x}^{2}} - 1$
using differential formula
$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{\frac{\mathrm{du}}{\mathrm{dx}} . v - \frac{\mathrm{dv}}{\mathrm{dx}} . u}{v} ^ 2$

hence here $u = {e}^{x}$ and $v = {x}^{2}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(\frac{d}{\mathrm{dx}} \left({e}^{x}\right)\right) . {x}^{2} - \left(\frac{d}{\mathrm{dx}} \left({x}^{2}\right)\right) . {e}^{x}}{{x}^{2}} ^ 2 - \frac{d}{\mathrm{dx}} \left(1\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left({e}^{x}\right) . {x}^{2} - \left(2 x\right) . {e}^{x}}{{x}^{4}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{x} . {x}^{2} - 2 x . {e}^{x}}{{x}^{4}} = \frac{x {e}^{x} - 2 {e}^{x}}{x} ^ 3$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{x} \left(x - 2\right)}{x} ^ 3$

Sep 17, 2017

$y ' = \frac{{e}^{x} \left(x - 2\right)}{{x}^{3}}$

Explanation:

We have: $y = \frac{{e}^{x}}{{x}^{2}} - 1$

$R i g h t a r r o w y ' = \frac{d}{\mathrm{dx}} \left(\frac{{e}^{x}}{{x}^{2}}\right) - \frac{d}{\mathrm{dx}} \left(1\right)$

$R i g h t a r r o w y ' = \frac{d}{\mathrm{dx}} \left(\frac{{e}^{x}}{{x}^{2}}\right) - 0$

$R i g h t a r r o w y ' = \frac{d}{\mathrm{dx}} \left(\frac{{e}^{x}}{{x}^{2}}\right)$

$R i g h t a r r o w y ' = \frac{{x}^{2} \cdot \frac{d}{\mathrm{dx}} \left({e}^{x}\right) - {e}^{x} \cdot \frac{d}{\mathrm{dx}} \left({x}^{2}\right)}{{\left({x}^{2}\right)}^{2}}$

$R i g h t a r r o w y ' = \frac{{x}^{2} \cdot {e}^{x} - {e}^{x} \cdot 2 x}{{x}^{4}}$

$R i g h t a r r o w y ' = \frac{{e}^{x} \left({x}^{2} - 2 x\right)}{{x}^{4}}$

$R i g h t a r r o w y ' = \frac{x {e}^{x} \left(x - 2\right)}{{x}^{4}}$

$\therefore y ' = \frac{{e}^{x} \left(x - 2\right)}{{x}^{3}}$