做ex_2的时候，碰到一个求梯度公式，在此手推一波. 如下：

$\frac{\delta J(\theta)}{\delta\theta_j}=\frac {1}{m}\sum_{i=1}^m(h_\theta(x)^{(i)}-y^{(i)})x^{(i)}_j$

1. 前提：cost function $J(\theta)=\frac 1m\sum_{i=1}^m[-y^{(i)}log(h_\theta(x^{(i)}))-(1-y^{(i)})log(1-h_\theta(x^{(i)}))]$hypothesis
   $h_\theta(x^{(i)})=g(x{(i)}\theta)$Logistic function
   $g(x)=\frac 1{1+exp(-x)}, exp(x)=e^x$
2. 推导过程:
   $\frac{\delta J(\theta)}{\delta\theta_j}=-\frac1m(y^{(i)}\frac{h^{'}_\theta(x^{(i)})}{h_\theta(x^{(i)})}+(1-y^{(i)})\frac{-h^{'}_\theta(x^{(i)})}{1-h_\theta(x^{(i)})})$
   $=-\frac1m\frac{y^{(i)}h^{'}_\theta(x^{(i)}(1-h_\theta(x^{(i)}))-(1-y^{(i)}){h^{'}_\theta(x^{(i)})}{h_\theta(x^{(i)})}}{h_\theta(x^{(i)})(1-h_\theta(x^{(i)}))}$
   $=-\frac1m\frac{(y^{(i)}-h_\theta(x^{(i)}))h^{'}_\theta(x^{(i)})}{h_\theta(x^{(i)})(1-h_\theta(x^{(i)}))}...............(1)$
   $设H(x)=x{(i)}\theta, 则h^{'}_\theta(x^{(i)})=-\frac{e^H(x)}{(1+e^-H(x))^2}H^{'}(x).......(2)$
   $h_\theta(x^{(i)})(1-h_\theta(x^{(i)}))= \frac{e^H(x)}{(1+e^-H(x))^2}..........(3)$
   $H^{'}(x)=\theta_j^{(i)}................(4)$
   $将(2)(3)(4)代入（1）得\frac{\delta J(\theta)}{\delta\theta_j}=\frac {1}{m}x^{(i)}_j(h_\theta(x)^{(i)}-y^{(i)})x^{(i)}_j$
   $$