I am going over the lectures on Machine Learning at Coursera.

I am struggling with the following. How can the partial derivative of


where $h_{\theta}(x)$ is defined as follows

$$h_{\theta}(x)=g(\theta^{T}x)$$ $$g(z)=\frac{1}{1+e^{-z}}$$

be $$ \frac{\partial}{\partial\theta_{j}}J(\theta) =\sum_{i=1}^{m}(h_\theta(x^{i})-y^i)x_j^i$$

In other words, how would we go about calculating the partial derivative with respect to $\theta$ of the cost function (the logs are natural logarithms):


  • $\begingroup$ I think to resolve $\theta$ by gradient will be hard way (or impossible??). Because it different with linear classfication, it will not has close form. So i suggest you can use other method example Newton's method. BTW, do you find $\theta$ using above way? $\endgroup$ – John Jul 22 '14 at 2:16
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    $\begingroup$ missing $\frac{1}{m}$ for the derivative of the Cost $\endgroup$ – bourneli Apr 20 '17 at 5:01

The reason is the following. We use the notation:

$$\theta x^i:=\theta_0+\theta_1 x^i_1+\dots+\theta_p x^i_p.$$


$$\log h_\theta(x^i)=\log\frac{1}{1+e^{-\theta x^i} }=-\log ( 1+e^{-\theta x^i} ),$$ $$\log(1- h_\theta(x^i))=\log(1-\frac{1}{1+e^{-\theta x^i} })=\log (e^{-\theta x^i} )-\log ( 1+e^{-\theta x^i} )=-\theta x^i-\log ( 1+e^{-\theta x^i} ),$$ [ this used: $ 1 = \frac{(1+e^{-\theta x^i})}{(1+e^{-\theta x^i})},$ the 1's in numerator cancel, then we used: $\log(x/y) = \log(x) - \log(y)$]

Since our original cost function is the form of:


Plugging in the two simplified expressions above, we obtain $$J(\theta)=-\frac{1}{m}\sum_{i=1}^m \left[-y^i(\log ( 1+e^{-\theta x^i})) + (1-y^i)(-\theta x^i-\log ( 1+e^{-\theta x^i} ))\right]$$, which can be simplified to: $$J(\theta)=-\frac{1}{m}\sum_{i=1}^m \left[y_i\theta x^i-\theta x^i-\log(1+e^{-\theta x^i})\right]=-\frac{1}{m}\sum_{i=1}^m \left[y_i\theta x^i-\log(1+e^{\theta x^i})\right],~~(*)$$

where the second equality follows from

$$-\theta x^i-\log(1+e^{-\theta x^i})= -\left[ \log e^{\theta x^i}+ \log(1+e^{-\theta x^i} ) \right]=-\log(1+e^{\theta x^i}). $$ [ we used $ \log(x) + \log(y) = log(x y) $ ]

All you need now is to compute the partial derivatives of $(*)$ w.r.t. $\theta_j$. As $$\frac{\partial}{\partial \theta_j}y_i\theta x^i=y_ix^i_j, $$ $$\frac{\partial}{\partial \theta_j}\log(1+e^{\theta x^i})=\frac{x^i_je^{\theta x^i}}{1+e^{\theta x^i}}=x^i_jh_\theta(x^i),$$

the thesis follows.

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    $\begingroup$ Can't upvote as I don't have 15 reputation just yet! :) Will google the maximum entropy principle as I have no clue what that is! as a side note I am not sure how you made the jump from log(1 - hypothesis(x)) to log(a) - log(b) but will raise another question for this as I don't think I can type latex here, really impressed with your answer! learning all this stuff on my own is proving to be quite a challenge thus the more kudos to you for providing such an elegant answer! :) $\endgroup$ – dreamwalker Aug 27 '13 at 13:54
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    $\begingroup$ yes!!! I couldn't see that you were using this property $\log(\frac{a}{b})=\log a-\log b$ Now everything makes sense :) Thank you so much! :) $\endgroup$ – dreamwalker Aug 27 '13 at 14:26
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    $\begingroup$ Awesome explanation, thank you very much! The only thing I am still struggling with is the very last line, how the derivative was made in $$\frac{\partial}{\partial \theta_j}\log(1+e^{\theta x^i})=\frac{x^i_je^{\theta x^i}}{1+e^{\theta x^i}}$$ ? Could you provide a hint for it? Thank you very much for the help! $\endgroup$ – Pedro Lopes Dec 1 '15 at 21:40
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    $\begingroup$ @codewarrior hope this helps. $$\frac{\partial}{\partial \theta_j}\log(1+e^{\theta x^i})=\frac{x^i_je^{\theta x^i}}{1+e^{\theta x^i}} $$ $$ = \frac{{x^i_j}}{{e^{-\theta x^i}*(1+e^{\theta x^i})}} $$ $$ =\frac{{x^i_j}}{{e^{-\theta x^i}+e^{-\theta x^i + \theta x^i}}} $$ $$ =\frac{{x^i_j}}{{e^{-\theta x^i}+e^{0}}} $$ $$ =\frac{{x^i_j}}{{e^{-\theta x^i}+1}} $$ $$ =\frac{{x^i_j}}{{1+e^{-\theta x^i}}} $$ $$ =x^i_j*h_\theta(x^i) $$ as $$ h_\theta(x^i) = \frac{{1}}{{1+e^{\theta x^i}}} $$ $\endgroup$ – Rudresha Parameshappa Jan 2 '17 at 13:06
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    $\begingroup$ @Israel, logarithm is usually base e in math. Take a look at When log is written without a base, is the equation normally referring to log base 10 or natural log? $\endgroup$ – gdrt Mar 11 '18 at 11:46

Pedro, => partial fractions

$$\log(1 - \frac{a}{b})$$

$$1 - \frac{a}{b} = \frac{b}{b} - \frac{a}{b} = \frac{b-a}{b},$$ $$\log(1 - \frac{a}{b}) = \log(\frac{b-a}{b}) = \log(b-a) - \log(b)$$


@pedro-lopes, it is called as: chain rule. $$(u(v))' = u(v)' * v'$$ For example: $$y = \sin(3x - 5)$$ $$u(v) = \sin(3x - 5)$$ $$v = (3x - 5)$$ $$y' = \sin(3x - 5)' = \cos(3x - 5) * (3 - 0) = 3\cos(3x-5)$$

Regarding: $$\frac{\partial}{\partial \theta_j}\log(1+e^{\theta x^i})=\frac{x^i_je^{\theta x^i}}{1+e^{\theta x^i}}$$ $$u(v) = \log(1+e^{\theta x^i})$$ $$v = 1+e^{\theta x^i}$$ $$\frac{\partial}{\partial \theta}\log(1+e^{\theta x^i}) = \frac{\partial}{\partial \theta}\log(1+e^{\theta x^i}) * \frac{\partial}{\partial \theta}(1+e^{\theta x^i}) = \frac{1}{1+e^{\theta x^i}} * (0 + xe^{\theta x^i}) = \frac{xe^{\theta x^i}}{1+e^{\theta x^i}} $$ Note that $$\log(x)' = \frac{1}{x}$$ Hope that I answered on your question!


We have, \begin{align*} L(\theta) &= -\frac{1}{m}\sum\limits_{i=1}^{m}{y_i. log P(y_i|x_i,\theta) + (1-y_i). \log{(1 - P(y_i|x_i,\theta))}} \\ h_\theta(x_i) &= P(y_i|x_i,\theta) = P(y_i=1|x_i,\theta) = \frac{1}{1+\exp{\left(-\sum\limits_k \theta_k x_i^k \right)}} \end{align*}

Then, \begin{align*} \log{(P(y_i|x_i,\theta))}=\log{(P(y_i=1|x_i,\theta))} &=-\log{\left(1+\exp{\left(-\sum\limits_k \theta_k x_i^k \right)} \right)} \\ \Rightarrow \frac{\partial }{\partial \theta_j} log P(y_i|x_i,\theta) =\frac{x_i^j.\exp{\left(-\sum\limits_k \theta_k x_i^k\right)}}{1+\exp{\left(-\sum\limits_k \theta_k x_i^k\right)}} &= x_i^j.\left(1-P(y_i|x_i,\theta)\right) \end{align*} and \begin{align*} \log{(1-P(y_i|x_i,\theta))}=\log{(1-P(y_i=1|x_i,\theta))} &=-\sum\limits_k \theta_k x_i^k -\log{\left(1+\exp{\left(-\sum\limits_k \theta_k x_i^k \right)} \right)} \\ \Rightarrow \frac{\partial }{\partial \theta_j} \log{(1 - P(y_i|x_i,\theta))} &= -x_i^j + x_i^j.\left(1-P(y_i|x_i,\theta)\right) = -x_i^j.P(y_i|x_i,\theta) \\ \end{align*}


\begin{align*} \frac{\partial }{\partial \theta_j} L(\theta) &= -\frac{1}{m}\sum\limits_{i=1}^{m}{y_i.\frac{\partial }{\partial \theta_j} log P(y_i|x_i,\theta) + (1-y_i).\frac{\partial }{\partial \theta_j} \log{(1 - P(y_i|x_i,\theta))}} \\ &=-\frac{1}{m}\sum\limits_{i=1}^{m}{y_i.x_i^j.\left(1-P(y_i|x_i,\theta)\right) - (1-y_i).x_i^j.P(y_i|x_i,\theta)} \\ &=-\frac{1}{m}\sum\limits_{i=1}^{m}{y_i.x_i^j - x_i^j.P(y_i|x_i,\theta)} \\ &=\frac{1}{m}\sum\limits_{i=1}^{m}{(P(y_i|x_i,\theta)-y_i).x_i^j} \end{align*} (Proved)


$${ J(\theta)=-\frac{1}{m} \sum_{i=1}^{m} y^i\log(h_\theta(x^i))+(1-y^i)\log(1-h_\theta(x^i)) }$$ where $h_\theta(x)$ is defined as follows $${ h_\theta(x)=g(\theta^Tx), }$$ $${ g(z)=\frac{1}{1+e^{-z}} }$$ Note that $g(z)'=g(z)*(1-g(z))$ and we can simply write right side of summation as $${ y\log(g)+(1-y)\log(1-g) }$$ and the derivative of it as $${ y \frac{1}{g}g'+(1-y) \left( \frac{1}{1-g}\right) (-g') \\ =\left( \frac{y}{g}- \frac{1-y}{1-g}\right) g' \\ = \frac{y(1-g)-g(1-y)}{g(1-g)}g' \\ = \frac{y-y*g-g+g*y}{g(1-g)}g' \\ = \frac{y-y*g-g+g*y}{g(1-g)}g(1-g)*x \\ =(y-g)*x }$$

and then we can rewrite above as $${ \frac{\partial}{\partial\theta_{j}}J(\theta) =\frac{1}{m}\sum_{i=1}^{m}(h_\theta(x^{i})-y^i)x_j^i }$$


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