Derivative of e^x/(1+e^x): Answer, Proof, Mistakes

Quotient rule on the logistic function

Because $e^{x}$ is its own derivative and the $1$ underneath contributes nothing, the quotient rule asks for $e^{x}\left(1+e^{x}\right)$ minus $e^{x}\cdot e^{x}$. Both of those products carry an $e^{2x}$, and that is the term that disappears.

Expand the top and it reads $e^{x}+e^{2x}-e^{2x}$. The two $e^{2x}$ terms are identical and opposite, so only $e^{x}$ is left standing.

Why the derivative equals f(1 - f)

Split the answer on purpose instead of leaving it as one fraction. Since $1-\frac{e^{x}}{1+e^{x}} = \frac{1}{1+e^{x}}$, the second factor below is exactly $1-f(x)$.

That is the logistic differential equation $\frac{dy}{dx} = y(1-y)$ from Unit 7, with carrying capacity $1$. The function on this page is one of its solutions, so the growth model and the quotient rule arrive at the same expression from opposite directions.

The graph follows from the sign. Both $e^{x}$ and $\left(1+e^{x}\right)^{2}$ are positive, so $f$ increases on the whole real line. The steepest point is $x=0$, where $f=\frac{1}{2}$ and $f'=\frac{1}{4}$: the inflection point sits at half the carrying capacity, as it does for every logistic curve.

The mistakes students make

All three are quotient rule slips: one divides instead of subtracting, one forgets to square the denominator, and one drops the subtracted product.

• Differentiating top and bottom separately and answering $\frac{e^{x}}{e^{x}} = 1$. The quotient rule subtracts two products, it does not divide two derivatives.
• Leaving the denominator unsquared and answering $\frac{e^{x}}{1+e^{x}}$. That is the original function, not its derivative.
• Forgetting to subtract the second product at all and answering $\frac{e^{x}+e^{2x}}{\left(1+e^{x}\right)^{2}}$. Expanding $e^{x}\left(1+e^{x}\right)$ gives $e^{x}+e^{2x}$, and the subtracted product $e^{x}\cdot e^{x}$ removes that $e^{2x}$ in full.