y = \operatorname{sech}^2 x

Graph of the Function $y = \operatorname{sech}^2 x$

$y = \operatorname{sech}^2 x$ is the hyperbolic secant squared. Since the derivative of $\tanh x$ is exactly this, it appears everywhere as the bell shape that emerges when a sigmoid is differentiated.

Domain and range

The domain is all real numbers
The range is $0 < y \leq 1$
The maximum $1$ is attained at $x = 0$
It is an even function

From $\cosh x \geq 1$ we get $0 < \operatorname{sech} x \leq 1$ , and squaring keeps the values in the same range.

The derivative of the hyperbolic tangent

Differentiating $\tanh x = \dfrac{\sinh x}{\cosh x}$ with the quotient rule produces this function.

\begin{align*} \frac{d}{dx}\tanh x &= \frac{\cosh^2 x - \sinh^2 x}{\cosh^2 x} \\ &= \frac{1}{\cosh^2 x} = \operatorname{sech}^2 x \end{align*}

So this function is the slope of $\tanh x$ itself. The way $\tanh$ rises most steeply near the origin and flattens at both ends appears here directly as the shape of the hump.

Monotonicity and extremum

The derivative is $y' = -2\operatorname{sech}^2 x \tanh x$ . Since $\operatorname{sech}^2 x$ is positive, the sign comes from $-\tanh x$ : the function increases for $x < 0$ and decreases for $x > 0$ . The maximum, both local and global, is at $(0, 1)$ .

Inflection points

Using $\operatorname{sech}^2 x = 1 - \tanh^2 x$ , the second derivative can be written as follows.

y'' = (6\tanh^2 x - 2)\operatorname{sech}^2 x

The sign changes when $\tanh^2 x = \dfrac{1}{3}$ , that is at $x = \pm\operatorname{artanh}\dfrac{1}{\sqrt{3}} \approx \pm 0.658$ . The value there is $1 - \dfrac{1}{3} = \dfrac{2}{3}$ , a tidy fraction for the height of an inflection point. The curve is concave down between them and concave up outside.

$x$	$y$	Position
$0$	$1$	maximum
$\approx \pm 0.658$	$\dfrac{2}{3}$	inflection point
$\pm 1$	$\approx 0.4200$	the skirt of the hump
$\pm 2$	$\approx 0.0707$	all but flat

Decay

For large $|x|$ we have $\operatorname{sech} x \approx 2e^{-|x|}$ , so the decay is exponential.

y \approx 4e^{-2|x|}

The $x$ -axis is a horizontal asymptote. The fall is not as steep as a Gaussian, but it is quick enough.

Integral and a probability density

The antiderivative is $\tanh x$ itself, so the integral over the whole line is as follows.

\int_{-\infty}^{\infty} \operatorname{sech}^2 x\,dx = \tanh(\infty) - \tanh(-\infty) = 2

Hence $\dfrac{1}{2}\operatorname{sech}^2 x$ is a probability density. Indeed $\dfrac{1 + \tanh x}{2} = \dfrac{1}{1 + e^{-2x}}$ , so it is the density of the distribution whose cumulative function is a logistic function.

Solitons

The KdV equation, which describes waves on shallow water, has the following travelling-wave solution¹.

u(x, t) = \frac{c}{2}\operatorname{sech}^2\frac{\sqrt{c}}{2}(x - ct)

This solitary wave, which advances at a constant speed without changing shape, is a soliton, and its cross-section is exactly this graph. The formula also shows that a faster soliton is taller and narrower. It is the theoretical form of the wave that would not change, observed by Russell on a canal in $1834$ .

Applications

In quantum mechanics a potential of the form $-\operatorname{sech}^2 x$ can be solved exactly and has the special property of reflecting no waves at all. In machine learning, the gradient of a network using $\tanh$ as its activation function is this very function, so one can see with the eye why the gradient falls exponentially for large inputs, the phenomenon known as the vanishing gradient².

Soliton, Wikipedia
Vanishing gradient problem, Wikipedia

Graph of the Function y=sech⁡2xy = \operatorname{sech}^2 xy=sech2x