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

Two routes to the same derivative

The first route rewrites the function as $\left(1+e^{x}\right)^{-1}$ and uses the chain rule. The outer derivative is $-\left(1+e^{x}\right)^{-2}$ and the inner derivative is $e^{x}$, since the $1$ contributes nothing.

The second route uses the quotient rule with $u=1$ and $v=1+e^{x}$. Because $u'=0$, the first term of $\frac{u'v-uv'}{v^{2}}$ vanishes and only the second survives.

Sign, shape, and the partner function

Both $e^{x}$ and $\left(1+e^{x}\right)^{2}$ are positive for every real $x$, so the derivative is negative everywhere and the function is strictly decreasing. It falls from $1$ as $x\to-\infty$ towards $0$ as $x\to\infty$, passing through $\frac{1}{2}$ at $x=0$, where the slope is $-\frac{1}{4}$.

The partner function $\frac{e^{x}}{1+e^{x}}$ is whatever is left over from $1$. Differentiating that identity forces the two derivatives to be negatives of each other, so the partner has derivative $+\frac{e^{x}}{\left(1+e^{x}\right)^{2}}$ and increases. Same expression, opposite sign.

One compact form is worth remembering: if $f(x)=\frac{1}{1+e^{x}}$ then $f'=-f\left(1-f\right)$, since $1-f$ is exactly the partner. Logistic models in the differential equations unit rest on this relationship between a function and its own rate of change.

The mistakes students make

The sign and the missing inner derivative account for most of the lost marks.

• Answering $\frac{e^{x}}{\left(1+e^{x}\right)^{2}}$ with no minus sign. That is the derivative of the partner function $\frac{e^{x}}{1+e^{x}}$, which increases, not of this one, which decreases.
• Answering $-\frac{1}{\left(1+e^{x}\right)^{2}}$, differentiating the outer power and forgetting the inner derivative $e^{x}$.
• Answering $0$ by differentiating the top and bottom separately, on the grounds that the numerator is constant. The derivative of a quotient is not the quotient of the derivatives.