From f380174723475cce5c921676daf26793323a1b8f Mon Sep 17 00:00:00 2001
From: Talia <31782251+TEChopra1000@users.noreply.github.com>
Date: Tue, 7 Jan 2020 15:00:15 -0800
Subject: [PATCH] Fixing autograd images

When being build, auto-grad links are not behaving as expected. Adding full links to fix images on this page.
---
 .../python/tutorials/packages/autograd/index.md           | 8 ++++----
 1 file changed, 4 insertions(+), 4 deletions(-)

diff --git a/docs/python_docs/python/tutorials/packages/autograd/index.md b/docs/python_docs/python/tutorials/packages/autograd/index.md
index fa321f72b839..155fb4ff4118 100644
--- a/docs/python_docs/python/tutorials/packages/autograd/index.md
+++ b/docs/python_docs/python/tutorials/packages/autograd/index.md
@@ -23,7 +23,7 @@
 
 Gradients are fundamental to the process of training neural networks, and tell us how to change the parameters of the network to improve its performance.
 
-![auto-gradient](/api/python/docs/_static/autograd_images/autograd_gradient.png)
+![auto-gradient](http://mxnet.incubator.apache.org/api/python/docs/_static/autograd_images/autograd_gradient.png)
 
 ### Long Answer:
 
@@ -39,7 +39,7 @@ Assuming we've calculated the gradient of each parameter with respect to the los
 
 We differentiate. [MXNet Gluon](/api/python/docs/tutorials/packages/gluon/index.html) uses Reverse Mode Automatic Differentiation (`autograd`) to backprogate gradients from the loss metric to the network parameters.
 
-![forward-backward](/api/python/docs/_static/autograd_images/autograd_forward_backward.png)
+![forward-backward](http://mxnet.incubator.apache.org/api/python/docs/_static/autograd_images/autograd_forward_backward.png)
 
 ### Long Answer:
 
@@ -217,7 +217,7 @@ def f(x):
 
 We can plot the resultant function for $x$ between 0 and 1, and we should recognise certain functions in segments of $x$. Starting with a quadratic curve from 0 to 1/2, we have a cubic curve from 1/2 to 2/3, a quartic from 2/3 to 3/4 and finally a flatline.
 
-![control-flow](/api/python/docs/_static/autograd_images/autograd_control_flow.png)
+![control-flow](https://mxnet.incubator.apache.org/api/python/docs/_static/autograd_images/autograd_control_flow.png)
 
 Using `autograd`, let's now find the gradient of this arbritrary function. We don't have a vectorized function in this case, because of the control flow, so let's also create a function to calculate the gradient using `autograd`.
 
@@ -234,7 +234,7 @@ grads = [get_grad(f, x).asscalar() for x in xs]
 print(grads)
 ```
 
-![flow-grad](/api/python/docs/_static/autograd_images/autograd_control_flow_grad.png)
+![flow-grad](https://mxnet.incubator.apache.org/api/python/docs/_static/autograd_images/autograd_control_flow_grad.png)
 
 We can calculate the gradients by hand in this situation (since it's a toy example), and for the four segments discussed before we'd expect $2x$, $3x^2$, $4x^3$ and 0. As a spot check, for $x=0.6$ the hand calculated gradient would be $3x^2=1.08$, which equals `1.08` as computed by `autograd`.