Derivatives of Activation functions

August 31, 2017

Here is my attempt at finding the derivatives of common activation functions of neural network namely sigmoid, tanh, ReLU and leaky ReLU functions.

1. Sigmoid function

$$ f(x) = {1 \over \sigma(x)} = 1 + e ^ {-x} $$

Differentiating both sides,

$$ f'(x) = {d \over dx} \biggl( { 1 \over \sigma(x)} \biggr) = {d \over dx} \bigl( 1 + e ^ {-x} \bigr) $$

$$ - {\sigma'(x) \over \sigma(x) ^ 2} = -e ^ {-x} $$

$$ - {\sigma'(x) \over \sigma(x) ^ 2} = 1 - f(x) $$

$$ - {\sigma'(x) \over \sigma(x) ^ 2} = 1 - f(x) $$

$$ - {\sigma'(x) \over \sigma(x) ^ 2} = 1 - {1 \over \sigma(x)} $$

$$ - {\sigma'(x) \over \sigma(x) ^ 2} = {\sigma(x) - 1 \over \sigma(x)} $$

Rearranging the terms,

$$ {\sigma'(x) \over \sigma(x) ^ 2} = {1 - \sigma(x) \over \sigma(x)} $$

$$ \sigma'(x) = \sigma(x) ^ 2 \biggl( {1 - \sigma(x) \over \sigma(x)} \biggr) $$

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


2. tanh function

$$ f(x) = tanh(x) = {e^x - e^{-x} \over e^x + e^{-x}} $$

Differentiating both sides,

$$ {d \over dx} tanh(x) = {d \over dx} \biggl( {sinh(x) \over cosh(x)} \biggr) $$

Using quotient rule,

$$ tanh'(x) = ( cosh(x) {d \over dx} sinh(x) - sinh(x) {d \over dx} cosh(x) ) / cosh^2(x) $$

$$ tanh'(x) = ( cosh(x)cosh(x) - sinh(x)sinh(x) ) / cosh^2(x) $$

$$ \boxed { tanh'(x) = 1 - tanh^2(x) } $$


3. ReLU function

$$ f(x) = max(0, x) $$

$$ f'(x) = \begin{cases} 1, & \text{if $x$ > 0} \ 0, & \text{otherwise} \end{cases} $$

Note: In software, we can use f'(x) = 1 for x = 0 (Prof. Andrew NG in Deep Learning Coursera Course)


4. Leaky ReLU function

$$ f(x) = max(0.01, x) $$

$$ f'(x) = \begin{cases} 1, & \text{if x > 0} \ 0.01, & \text{otherwise} \end{cases} $$

Note: In software, we can use f'(x) = 1 or 0.01 for x = 0 (Prof. Andrew NG in Deep Learning Coursera Course)

Ganessh Kumar R P © 2020 • All rights reserved.