I'm trying to calculate the gradient of the following multi-variable function:
$$ f(m,b) = \frac{1}{n}\sum_{i=0}^n {(y_i-(mx_i + b))^2}$$
I want to use Chain rule as $${(y_i-(mx_i + b))^2}$$ is a composite function. I'd like to calculate $$\frac{df}{dm}$$ and $$\frac{df}{db}$$
Can somebody show me how to derive this? Thank you.
Let's consider the i-th component of the sum:
$$f_i(m,b)=\frac{1}{n}{(y_i-(mx_i + b))^2}$$
thus:
$$\frac{\partial f_i}{\partial m}=\frac{2}{n}{(y_i-(mx_i + b))}(-x_i)$$
$$\frac{\partial f_i}{\partial b}=\frac{2}{n}{(y_i-(mx_i + b))}(-1)$$
and then adding up.
$$\frac{\partial f}{\partial m}= \frac{2}{n}\sum_{i=0}^n {-x_i(y_i-(mx_i + b))}$$
$$\frac{\partial f}{\partial b}= \frac{2}{n}\sum_{i=0}^n {-(y_i-(mx_i + b))}$$
NOTE
The functions to derive are in the form:
$$f(t)=[g(t)]^2\implies f'(t)=2g(t)g'(t)$$