What Is the Derivative of the Sigmoid Function?

The derivative of the logistic sigmoid has a particularly simple form: it equals the sigmoid’s value multiplied by one minus that value. This identity is useful for differentiating sigmoid expressions and for computing gradients in models such as neural networks.

The logistic sigmoid

The logistic sigmoid is the function

$$\sigma(x)=\frac{1}{1+e^{-x}}.$$

It maps every real input to a value strictly between 0 and 1. In some fields, “sigmoid” is used more broadly for any S-shaped function, but this logistic function is the standard meaning in many calculus and machine-learning contexts.

Deriving the derivative

Write the sigmoid as a power of its denominator and apply the chain rule:

$$\begin{aligned} \sigma'(x) &=\frac{d}{dx}(1+e^{-x})^{-1}\\ &=-(1+e^{-x})^{-2}(-e^{-x})\\ &=\frac{e^{-x}}{(1+e^{-x})^2}. \end{aligned}$$

This is already a correct derivative. To express it in terms of the sigmoid itself, note that

$$1-\sigma(x)=1-\frac{1}{1+e^{-x}}=\frac{e^{-x}}{1+e^{-x}}.$$

Multiplying this by $\sigma(x)$ gives the derivative:

$$\boxed{\sigma'(x)=\sigma(x)\bigl(1-\sigma(x)\bigr)}.$$

The two derivative forms are equivalent. Use the exponential form when you want a formula directly in terms of $x$, or the compact form when you already know the sigmoid’s output.

Applying the chain rule

If the sigmoid takes an inner function $g(x)$ as its input, differentiate both the outer sigmoid and the inner function:

$$\frac{d}{dx}\sigma(g(x))=\sigma(g(x))\bigl(1-\sigma(g(x))\bigr)g'(x).$$

For example, if $f(x)=\sigma(3x+2)$, then the inner function has derivative $3$, so

$$f'(x)=3\sigma(3x+2)\bigl(1-\sigma(3x+2)\bigr).$$

More generally, for $f(x)=\sigma(ax+b)$, the derivative is $a\sigma(ax+b)(1-\sigma(ax+b))$. The factor $a$ is essential: it comes from differentiating the input, not from the sigmoid itself.

For a neural-network unit with $z=\mathbf{w}^{\mathsf T}\mathbf{x}+b$ and output $y=\sigma(z)$, the derivative with respect to $z$ is $y(1-y)$. By the chain rule, the gradient with respect to the input vector is $y(1-y)\mathbf{w}$.

Worked example

Take $x=\ln 3$. Since $e^{-\ln 3}=1/3$, the sigmoid value is

$$\sigma(\ln 3)=\frac{1}{1+1/3}=\frac{3}{4}.$$

Using the compact derivative formula,

$$\sigma'(\ln 3)=\frac{3}{4}\left(1-\frac{3}{4}\right)=\frac{3}{16}.$$

The same result follows from the exponential form:

$$\frac{e^{-\ln 3}}{(1+e^{-\ln 3})^2} =\frac{1/3}{(4/3)^2} =\frac{3}{16}.$$

If instead $h(t)=\sigma(2t)$ and $t=(\ln 3)/2$, then the inner input is $2t=\ln 3$. The chain rule gives $h'(t)=2\sigma(2t)(1-\sigma(2t))$, so at this value of $t$, $h'(t)=3/8$.

What the derivative tells you

The derivative is always positive, so the logistic sigmoid is increasing. It reaches its largest slope at $x=0$: there, $\sigma(0)=1/2$, and therefore $\sigma'(0)=1/4$. As $x$ moves far in either direction, the sigmoid approaches 0 or 1 and its derivative approaches 0. This flattening at the ends is often called saturation.

Common mistakes

• Dropping the chain-rule factor: The derivative of $\sigma(3x+2)$ includes a factor of $3$; the derivative of $\sigma(x)$ does not.
• Using the wrong sign: Differentiating $e^{-x}$ gives $-e^{-x}$. The negative from differentiating the reciprocal form cancels it, leaving a positive derivative.
• Confusing the derivative with the sigmoid itself: The derivative is $\sigma(x)(1-\sigma(x))$, not just $1-\sigma(x)$.
• Assuming every sigmoid has this formula: The identity applies to the logistic sigmoid. For example, if the function is defined as $\tanh(x)$, its derivative is $1-\tanh^2(x)$ instead.