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
concepts that are helpful:
- 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:
$$ \frac{\partial}{\partial \theta_1} f(\theta_0, \theta_1)^{(i)} = 0 + (\theta_{1})^1
x^{(i)} - 0 = 1 \times \theta_1^{(1-1=0)} x^{(i)} = 1 \times 1 \times x^{(i)} =
x^{(i)} \tag{11}$$
Thus, the entire answer becomes:
$$ \frac{\partial}{\partial \theta_1} g(f(\theta_0, \theta_1)^{(i)}) =
\frac{\partial}{\partial \theta_1} g(\theta_0, \theta_1) \frac{\partial}{\partial
\theta_1} f(\theta_0, \theta_1)^{(i)} = \tag{12}$$
$$\frac{1}{m} \sum_{i=1}^m f(\theta_0, \theta_1)^{(i)} \frac{\partial}{\partial
\theta_1}f(\theta_0, \theta_1)^{(i)} = \frac{1}{m} \sum_{i=1}^m \left(\theta_0 +
\theta_{1}x^{(i)} - y^{(i)}\right) x^{(i)}$$
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.
Best Answer
The idea is that you do not need to do this if you have the gradient. The gradient exactly tells you which direction is the "best" (i.e., leading to fastest increase or decrease) around any point you evaluate it at.
In fact, your "parameter tweaking" is simply the finite difference approximation of the derivative. In other words, "tweaking" the parameters in each direction is simply an approximation of the gradient anyway (or equivalently the gradient is a finite differencing with an infinitely small step size). So it is doing what you think it is doing.
One thing you might wonder is how this can work over long distances. For instance, imagine a function with many "hills" - we won't be able to "see" far enough (because the gradient only looks very close by) to find our way over the hills and valleys. This is true. It is what people mean when they say that gradient descent methods get stuck in local minima. It may not find the global one.
As Wikipedia says:
Note that it also only works for optimizing differentiable functions for local extrema. A very similar approach to the one you proposed that works in the discrete case is called hill climbing.