逻辑回归求梯度的数学推导

做ex_2的时候，碰到一个求梯度公式，在此手推一波. 如下：

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

1. 前提：cost function$$J(\theta )=\frac{1}{m}\sum _{i=1}^m[-y^{(i)}log(h_{\theta }(x^{(i)}))-(1-y^{(i)})log(1-h_{\theta }(x^{(i)}))]$$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:
   $$\frac{\delta J(\theta )}{\delta {\theta }_j}=-\frac{1}{m}(y^{(i)}\frac{h_{\theta }^{{}^{\prime }}(x^{(i)})}{h_{\theta }(x^{(i)})}+(1-y^{(i)})\frac{-h_{\theta }^{{}^{\prime }}(x^{(i)})}{1-h_{\theta }(x^{(i)})})$$
   $$=-\frac{1}{m}\frac{y^{(i)}{h}_{\theta }^{{}^{\prime }}(x^{(i)}(1-h_{\theta }(x^{(i)}))-(1-y^{(i)})h_{\theta }^{{}^{\prime }}(x^{(i)})h_{\theta }(x^{(i)})}{h_{\theta }(x^{(i)})(1-h_{\theta }(x^{(i)}))}$$
   $$=-\frac{1}{m}\frac{(y^{(i)}-h_{\theta }(x^{(i)}))h_{\theta }^{{}^{\prime }}(x^{(i)})}{h_{\theta }(x^{(i)})(1-h_{\theta }(x^{(i)}))}...............(1)$$
   $$设H(x)=x(i)\theta ,则h_{\theta }^{{}^{\prime }}(x^{(i)})=-\frac{e^H(x)}{(1+e^{-}H(x){)}^2}{H}^{{}^{\prime }}(x).......(2)$$
   $$h_{\theta }(x^{(i)})(1-h_{\theta }(x^{(i)}))=\frac{e^H(x)}{(1+e^{-}H(x){)}^2}..........(3)$$
   $$H^{{}^{\prime }}(x)={\theta }_j^{(i)}................(4)$$
   $$将(2)(3)(4)代入（1）得\frac{\delta J(\theta )}{\delta {\theta }_j}=\frac{1}{m}{x}_j^{(i)}(h_{\theta }(x{)}^{(i)}-y^{(i)})x_j^{(i)}$$