## Issue

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.)

Thanks!

## Solution

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 **