Is tensoflow pow(x, y) differentiable by y?


Some of the tensorflow functions are only differentiable by some of the arguments.

I want to know if tf.pow(x, y) is differentiable by y? (I am pretty sure it is differentiable by x.)



Yes, tf.pow(x, y) is differentiable by y in tensorflow.

You can test the following scenario.

x = tf.Variable(3.0)
y = tf.Variable(3.0)

with tf.GradientTape() as tape:
  z = tf.pow(x, y)

dz_dy = tape.gradient(z, y)

dz/dy is (x^y)*log(x) so we should get a value of (3^3)*(log(3)) which is exactly what we see with tf

dz_dy.numpy() = 29.662533

Answered By – Max Feinberg

This Answer collected from stackoverflow, is licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0

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