$\ln x$ 导数的推导
$$ \frac{d}{dx}\ln x = \lim_{h \to 0} \frac{\ln(x + h) - \ln(x)}{h} = \lim_{h \to 0} \frac{\ln\left(\frac{x + h}{x}\right)}{h} = \lim_{h \to 0} \frac{\ln\left(1 + \frac{h}{x}\right)}{h} $$
$$ = \lim_{h \to 0} \frac{\ln(x + h) - \ln(x)}{x \cdot \left(\frac{h}{x}\right)} = \frac{1}{x} \cdot \lim_{h \to 0} \frac{\ln\left(1 + \left(\frac{h}{x}\right)\right)}{\frac{h}{x}} = \frac{1}{x} \cdot \lim_{h \to 0} \ln\left(1 + \frac{h}{x}\right)^{\frac{x}{h}} $$
因为：
$$ \lim_{u \to 0}\left(1 + u\right)^{\frac{1}{u}} = e $$
则有：
$$ \frac{d}{dx}\ln x = \frac{1}{x} \cdot \ln\left(\lim_{\frac{h}{x} \to 0} \left(1 + \frac{h}{x}\right)^{\frac{x}{h}}\right) = \frac{1}{x} \cdot \ln e = \frac{1}{x} $$
逻辑回归损失函数导数
$$ J(\theta) = \frac{1}{m}\sum_{i=1}^m Cost\left(h_{\theta}\left(x^{(i)}\right),y^{(i)}\right) = -\frac{1}{m}[\sum_{i=1}^m y^{(i)}\log h_{\theta}\left(x^{(i)}\right) + \left(1 - y^{(i)}\right)\log\left(1 - h_{\theta}\left(x^{(i)}\right)\right)] $$
因为：$ y = \frac{1}{1 + e^{-x}} $, $(1 - y) = \frac{e^{-x}}{1 + e^{-x}} $ 可知 $ \frac{dy}{dx} = y \cdot (1 - y) = \frac{e^{-x}}{(1 + e^{-x})^2} $， 则有：
$$ \frac{\partial}{\partial\theta_j}J(\theta) = -\frac{1}{m}[\sum_{i=1}^m y^{(i)} \frac{h_{\theta}^{'}\left(x^{(i)}\right)}{h_{\theta}\left(x^{(i)}\right)}\cdot x_j^{(i)} - \left(1 - y^{(i)}\right)\frac{h_{\theta}^{'}\left(x^{(i)}\right)}{1 - h_{\theta}\left(x^{(i)}\right)}\cdot x_j^{(i)}] $$
$$ = -\frac{1}{m}\sum_{i=1}^m \left(y^{(i)}\left(1 - h_{\theta}(x^{(i)})\right) - \left(1 - y^{(i)}\right)h_{\theta}(x^{(i)})\right) x_j^{(i)} $$
$$ = -\frac{1}{m}\sum_{i=1}^m \left(y^{(i)} - h_{\theta}(x^{(i)})\right)x_j^{(i)} = \frac{1}{m}\sum_{i=1}^m \left(h_{\theta}(x^{(i)}) - y^{(i)}\right)x_j^{(i)} $$