Softsign Function Derivative Calculator

Results

Softsign First Derivative

0.4444444444

phi'(x)

phi(x) = x / (1 + \|x\|)	0.3333333333
phi'(x) = 1 / (1 + \|x\|)^2	0.4444444444

What is the Softsign function?

The Softsign function is a smooth activation function used in neural networks, defined as $\phi(x) = x / (1 + |x|)$. It maps any real input into the open range (-1, 1), much like the hyperbolic tangent, but approaches its saturation limits more slowly. This gentler saturation can help reduce the vanishing-gradient problem during training. This calculator returns both the function value $\phi(x)$ and, as its primary output, the first derivative $\phi'(x)$.

Two overlaid curves on x-y axes: an S-shaped curve flattening toward horizontal asymptotes and a bell-shaped curve peaking at the center — The Softsign function $\phi(x) = x/(1+|x|)$ (S-shaped) and its derivative $\phi'(x) = 1/(1+|x|)^2$ (bell-shaped).

How to use this calculator

Enter any real number for x — positive, negative, or zero — and read off $\phi'(x)$ (the slope of the Softsign curve) along with $\phi(x)$ (the activation output). No units are involved; x is a pure dimensionless real number. The default input is $x = 0.5$.

The formula explained

The derivative of $\phi(x) = x / (1 + |x|)$ is

$$\phi'(x) = \frac{1}{\left(1 + |x|\right)^{2}}$$

Because the denominator $(1 + |x|)$ is always at least 1, the derivative is always strictly positive and lies in the range (0, 1]. Its maximum value of 1 occurs at $x = 0$, where the function is steepest. As $|x|$ grows large, $\phi'(x)$ shrinks toward 0, reflecting the function's saturation.

Single bell-shaped curve peaking at value 1 above the origin, symmetric and decaying to zero on both sides — Graph of the derivative $\phi'(x) = 1/(1+|x|)^2$, symmetric about $x=0$ with a maximum of 1 at the origin.

Worked example

For $x = 0.5$: $|x| = 0.5$, so $1 + |x| = 1.5$. The function value is

$$\phi(0.5) = \frac{0.5}{1.5} = 0.333333\ldots$$

and the derivative is

$$\phi'(0.5) = \frac{1}{1.5^{2}} = \frac{1}{2.25} = 0.444444\ldots$$

So the Softsign curve at $x = 0.5$ has an output near 0.3333 and a slope near 0.4444.

FAQ

Is the Softsign differentiable everywhere? Yes. Even though $|x|$ has a kink at $x = 0$, the left and right derivatives of $\phi$ both equal 1 there, so $\phi$ is differentiable at 0 and everywhere else.

Can the derivative ever be negative? No. $\phi'(x) = 1 / (1 + |x|)^2$ is always positive, since it is one divided by a squared positive number.

How does Softsign compare to tanh? Both saturate to (-1, 1), but Softsign uses polynomial $(1/x^2)$ tails rather than tanh's exponential tails, so it saturates more slowly and keeps slightly larger gradients far from zero.

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