How to compute the partial derivative

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Kumar Thapa
Kumar Thapa 2019 年 10 月 22 日
コメント済み: Kumar Thapa 2019 年 10 月 22 日
Compute the partial derivative of the following loss function with respect to θj:
J(θ0, θ1, θ2) = -1/m sum_{i = 1}^m [y(i) * log(hθ(x(i))) + (1 - y(i)) log(1 - hθ(x(i)))],
where hθ(x) = 1/(1 + exp(-(θ0 + θ1 x1 + θ2 x2)).
  2 件のコメント
KALYAN ACHARJYA
KALYAN ACHARJYA 2019 年 10 月 22 日
Define variables as syms and do diff respectively
Kumar Thapa
Kumar Thapa 2019 年 10 月 22 日
sysm x1 x2 theta0 theta1 theta2 y
h(theta0, theta1, theta2) = 1/(1+exp(-(theta0 + theta1*x1 + theta2*x2)));
I dont know how to differentiate it using this :
diff(function, var).

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Jon
Jon 2019 年 10 月 22 日
As @Kalyan suggested you could use the symbolic toolbox if you have it. Otherwise you could just go to the definition of a partial derivative. Evaluate your function to quantify how much it changes when you make small perturbations to the variable you are taking the derivative with respect to. Then the partial derivative is the small change in the function divided by the small change to the input variable. You have to be careful though on the size of the perturbation, too small and the round off causes problems, too big and it isn't a very good approximation. Maybe plot your estimate of your derivative versus perturbation size to get an idea of the behavior. Hopefully has a relatively stable plateau between too small and too big.

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