Jul 26, 2016

Please see the explanation section below.

#### Explanation:

If $y$ is a function of $x$ and $x$ a function of $t$, then the chain rule tells us that

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{\mathrm{dy}}{\mathrm{dx}} \frac{\mathrm{dx}}{\mathrm{dt}}$.

The problem here therefore amounts to showing that

$\frac{\mathrm{dx}}{\mathrm{dt}} = x \left(1 - x\right)$.

With $x = \frac{{e}^{t}}{1 + {e}^{t}}$, we use the quotient rule to get

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

$= {e}^{t} / {\left(1 + {e}^{t}\right)}^{2}$.

Substituting for $x$ in $x \left(1 - x\right)$, we get

$x \left(1 - x\right) = {e}^{t} / \left(1 + {e}^{t}\right) \left(1 - {e}^{t} / \left(1 + {e}^{t}\right)\right)$

$= {e}^{t} / \left(1 + {e}^{t}\right) \left(\frac{1 + {e}^{t} - {e}^{t}}{1 + {e}^{t}}\right)$

$= {e}^{t} / {\left(1 + {e}^{t}\right)}^{2}$.

We conclude that $\frac{\mathrm{dx}}{\mathrm{dt}} = x \left(1 - x\right)$ as required.