Solution 10 Ass3

### Solution 10:
$
L = -t \log(y) - (1 - t)\log(1 - y) \\
L = -t \log(\frac{1}{1+\exp(-a)}) - (1 - t)\log(1 - \frac{1}{1+\exp(-a)}) \\
L = -t \log(\frac{1}{1+\exp(-w^T x)}) - (1 - t)\log(1 - \frac{1}{1+\exp(-w^T x)}) \\
\frac{\partial L}{\partial w^T} = - t (\frac{x exp(xw^T)}{(1+exp(xw^T))^2}) - (1 - t) (\frac{x exp(xw^T)}{(1+exp(xw^T))^2}) \\
\frac{\partial L}{\partial w^T} = \frac{-t x exp(xw^T)}{(1+exp(xw^T))^2} - \frac{(1 - t) * x exp(xw^T)}{(1+exp(xw^T))^2} \\
\frac{\partial L}{\partial w^T} = \frac{-t x exp(xw^T)}{(1+exp(xw^T))^2} - \frac{x exp(xw^T) - t x exp(xw^T)}{(1+exp(xw^T))^2} \\
\frac{\partial L}{\partial w^T} = \frac{-t x exp(xw^T) - (exp(xw^T) - t x exp(xw^T))}{(1+exp(xw^T))^2} \\
\frac{\partial L}{\partial w^T} = \frac{-t x exp(xw^T) - exp(xw^T) + t x exp(xw^T)}{(1+exp(xw^T))^2} \\
\frac{\partial L}{\partial w^T} = \frac{- exp(xw^T)}{(1+exp(xw^T))^2} \\
$