# Partial derivative in gradient descent for two variables

I’ve started taking an online machine learning class, and the first learning algorithm that we are going to be using is a form of linear regression using gradient descent. I don’t have much of a background in high level math, but here is what I understand so far.

Given $$m$$ number of items in our learning set, with $$x$$ and $$y$$ values, we must find the best fit line $$h_\theta(x) = \theta_0+\theta_1x$$ . The cost function for any guess of $$\theta_0,\theta_1$$ can be computed as:

$$J(\theta_0,\theta_1) = \frac{1}{2m}\sum_{i=1}^m(h_\theta(x^{(i)}) – y^{(i)})^2$$

where $$x^{(i)}$$ and $$y^{(i)}$$ are the $$x$$ and $$y$$ values for the $$i^{th}$$ component in the learning set. If we substitute for $$h_\theta(x)$$,

$$J(\theta_0,\theta_1) = \frac{1}{2m}\sum_{i=1}^m(\theta_0 + \theta_1x^{(i)} – y^{(i)})^2$$

Then, the goal of gradient descent can be expressed as

$$\min_{\theta_0, \theta_1}\;J(\theta_0, \theta_1)$$

Finally, each step in the gradient descent can be described as:

$$\theta_j := \theta_j – \alpha\frac{\partial}{\partial\theta_j} J(\theta_0,\theta_1)$$

for $$j = 0$$ and $$j = 1$$ with $$\alpha$$ being a constant representing the rate of step.

I have no idea how to do the partial derivative. I have never taken calculus, but conceptually I understand what a derivative represents. The instructor gives us the partial derivatives for both $$\theta_0$$ and $$\theta_1$$ and says not to worry if we don’t know how it was derived. (I suppose, technically, it is a computer class, not a mathematics class) However, I would very much like to understand this if possible. Could someone show how the partial derivative could be taken, or link to some resource that I could use to learn more? I apologize if I haven’t used the correct terminology in my question; I’m very new to this subject.

The answer above is a good one, but I thought I’d add in some more “layman’s” terms that helped me better understand concepts of partial derivatives. The answers I’ve seen here and in the Coursera forums leave out talking about the chain rule, which is important to know if you’re going to get what this is doing…

It’s helpful for me to think of partial derivatives this way: the variable you’re
focusing on is treated as a variable, the other terms just numbers. Other key

• For “regular derivatives” of a simple form like $F(x) = cx^n$ , the derivative is simply $F'(x) = cn \times x^{n-1}$
• The derivative of a constant (a number) is 0.
• Summations are just passed on in derivatives; they don’t affect the derivative. Just copy them down in place as you derive.

Also, it should be mentioned that the chain
rule
is being used. The chain rule says
that (in clunky laymans terms), for $g(f(x))$, you take the derivative of $g(f(x))$,
treating $f(x)$ as the variable, and then multiply by the derivative of $f(x)$. For
our cost function, think of it this way:

$$g(\theta_0, \theta_1) = \frac{1}{2m} \sum_{i=1}^m \left(f(\theta_0, \theta_1)^{(i)}\right)^2 \tag{1}$$

$$f(\theta_0, \theta_1)^{(i)} = \theta_0 + \theta_{1}x^{(i)} – y^{(i)} \tag{2}$$

To show I’m not pulling funny business, sub in the definition of $f(\theta_0, \theta_1)^{(i)}$ into the definition of $g(\theta_0, \theta_1)$ and you get:

$$g(f(\theta_0, \theta_1)^{(i)}) = \frac{1}{2m} \sum_{i=1}^m \left(\theta_0 + \theta_{1}x^{(i)} – y^{(i)}\right)^2 \tag{3}$$

This is, indeed, our entire cost function.

Thus, the partial derivatives work like this:

$$\frac{\partial}{\partial \theta_0} g(\theta_0, \theta_1) = \frac{\partial}{\partial \theta_0} \frac{1}{2m} \sum_{i=1}^m \left(f(\theta_0, \theta_1)^{(i)}\right)^2 = 2 \times \frac{1}{2m} \sum_{i=1}^m \left(f(\theta_0, \theta_1)^{(i)}\right)^{2-1} = \tag{4}$$

$$\frac{1}{m} \sum_{i=1}^m f(\theta_0, \theta_1)^{(i)}$$

In other words, just treat $f(\theta_0, \theta_1)^{(i)}$ like a variable and you have a
simple derivative of $\frac{1}{2m} x^2 = \frac{1}{m}x$

$$\frac{\partial}{\partial \theta_0} f(\theta_0, \theta_1)^{(i)} = \frac{\partial}{\partial \theta_0} (\theta_0 + \theta_{1}x^{(i)} – y^{(i)}) \tag{5}$$

And $\theta_1, x$, and $y$ are just “a number” since we’re taking the derivative with
respect to $\theta_0$, so the partial of $g(\theta_0, \theta_1)$ becomes:

$$\frac{\partial}{\partial \theta_0} f(\theta_0, \theta_1) = \frac{\partial}{\partial \theta_0} (\theta_0 + [a \ number][a \ number]^{(i)} – [a \ number]^{(i)}) = \frac{\partial}{\partial \theta_0} \theta_0 = 1 \tag{6}$$

So, using the chain rule, we have:

$$\frac{\partial}{\partial \theta_0} g(f(\theta_0, \theta_1)^{(i)}) = \frac{\partial}{\partial \theta_0} g(\theta_0, \theta_1) \frac{\partial}{\partial \theta_0}f(\theta_0, \theta_1)^{(i)} \tag{7}$$

And subbing in the partials of $g(\theta_0, \theta_1)$ and $f(\theta_0, \theta_1)^{(i)}$
from above, we have:

$$\frac{1}{m} \sum_{i=1}^m f(\theta_0, \theta_1)^{(i)} \frac{\partial}{\partial \theta_0}f(\theta_0, \theta_1)^{(i)} = \frac{1}{m} \sum_{i=1}^m \left(\theta_0 + \theta_{1}x^{(i)} – y^{(i)}\right) \times 1 = \tag{8}$$

$$\frac{1}{m} \sum_{i=1}^m \left(\theta_0 + \theta_{1}x^{(i)} – y^{(i)}\right)$$

What about the derivative with respect to $\theta_1$?

Our term $g(\theta_0, \theta_1)$ is identical, so we just need to take the derivative
of $f(\theta_0, \theta_1)^{(i)}$, this time treating $\theta_1$ as the variable and the
other terms as “just a number.” That goes like this:

$$\frac{\partial}{\partial \theta_1} f(\theta_0, \theta_1)^{(i)} = \frac{\partial}{\partial \theta_1} (\theta_0 + \theta_{1}x^{(i)} – y^{(i)}) \tag{9}$$

$$\frac{\partial}{\partial \theta_1} f(\theta_0, \theta_1)^{(i)} = \frac{\partial}{\partial \theta_1} ([a \ number] + \theta_{1}[a \ number, x^{(i)}] – [a \ number]) \tag{10}$$

Note that the “just a number”, $x^{(i)}$, is important in this case because the
derivative of $c \times x$ (where $c$ is some number) is $\frac{d}{dx}(c \times x^1) = c \times 1 \times x^{(1-1=0)} = c \times 1 \times 1 = c$, so the number will carry
through. In this case that number is $x^{(i)}$ so we need to keep it. Thus, our
derivative is:

$$\frac{\partial}{\partial \theta_1} f(\theta_0, \theta_1)^{(i)} = 0 + (\theta_{1})^1 x^{(i)} – 0 = 1 \times \theta_1^{(1-1=0)} x^{(i)} = 1 \times 1 \times x^{(i)} = x^{(i)} \tag{11}$$

$$\frac{\partial}{\partial \theta_1} g(f(\theta_0, \theta_1)^{(i)}) = \frac{\partial}{\partial \theta_1} g(\theta_0, \theta_1) \frac{\partial}{\partial \theta_1} f(\theta_0, \theta_1)^{(i)} = \tag{12}$$

$$\frac{1}{m} \sum_{i=1}^m f(\theta_0, \theta_1)^{(i)} \frac{\partial}{\partial \theta_1}f(\theta_0, \theta_1)^{(i)} = \frac{1}{m} \sum_{i=1}^m \left(\theta_0 + \theta_{1}x^{(i)} – y^{(i)}\right) x^{(i)}$$

A quick addition per @Hugo’s comment below. Let’s ignore the fact that we’re dealing with vectors at all, which drops the summation and $fu^{(i)}$ bits. We can also more easily use real numbers this way.

$\require{cancel}$

Let’s say $x = 2$ and $y = 4$.

So, for part 1 you have:

$$\frac{\partial}{\partial \theta_0} (\theta_0 + \theta_{1}x – y)$$

Filling in the values for $x$ and $y$, we have:

$$\frac{\partial}{\partial \theta_0} (\theta_0 + 2\theta_{1} – 4)$$

We only care about $\theta_0$, so $\theta_1$ is treated like a constant (any number, so let’s just say it’s 6).

$$\frac{\partial}{\partial \theta_0} (\theta_0 + (2 \times 6) – 4) = \frac{\partial}{\partial \theta_0} (\theta_0 + \cancel8) = 1$$

Using the same values, let’s look at the $\theta_1$ case (same starting point with $x$ and $y$ values input):

$$\frac{\partial}{\partial \theta_1} (\theta_0 + 2\theta_{1} – 4)$$

In this case we do care about $\theta_1$, but $\theta_0$ is treated as a constant; we’ll do the same as above and use 6 for it’s value:

$$\frac{\partial}{\partial \theta_1} (6 + 2\theta_{1} – 4) = \frac{\partial}{\partial \theta_1} (2\theta_{1} + \cancel2) = 2 = x$$

The answer is 2 because we ended up with $2\theta_1$ and we had that because $x = 2$.

Hopefully the clarifies a bit on why in the first instance (wrt $\theta_0$) I wrote “just a number,” and in the second case (wrt $\theta_1$) I wrote “just a number, $x^{(i)}$. While it’s true that $x^{(i)}$ is still “just a number”, since it’s attached to the variable of interest in the second case it’s value will carry through which is why we end up at $x^{(i)}$ for the result.