# How do you find dy/dx if x=e^t/t and y=sqrte^t?

Jan 26, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = {t}^{2} / \left(2 \left(t - 1\right) \sqrt{{e}^{t}}\right)$

#### Explanation:

You can calculate the two differentials separately and then take the quotient:

$x = {e}^{t} / t \implies \mathrm{dx} = \frac{d}{\mathrm{dt}} \left({e}^{t} / t\right) \mathrm{dt} = \left({e}^{t} / t - {e}^{t} / {t}^{2}\right) \mathrm{dt} = {e}^{t} / {t}^{2} \left(t - 1\right) \mathrm{dt}$

$y = \sqrt{{e}^{t}} \implies \mathrm{dy} = \frac{d}{\mathrm{dt}} \left(\sqrt{{e}^{t}}\right) \mathrm{dt} = {e}^{t} / \left(2 \sqrt{{e}^{t}}\right) \mathrm{dt} = \frac{1}{2} \sqrt{{e}^{t}} \mathrm{dt}$

So:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \frac{\sqrt{{e}^{t}} \mathrm{dt}}{{e}^{t} / {t}^{2} \left(t - 1\right) \mathrm{dt}} = {t}^{2} / \left(2 \left(t - 1\right) \sqrt{{e}^{t}}\right)$