# derivative of an exponential function with respect to $\beta$ [closed]

If my $$f(x) = \sum_{i=1}^{n }x_i - \frac{\sum_{t_i\in(R_i)}x_ie^{x_i\beta}}{\sum_{t_i\in(R_i)}e^{x_i\beta}}$$

What is the first derivative of this function with respect to $$\beta$$ ?

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• Why do you have $\beta_p$ instead of $\beta$? – marty cohen Apr 24 at 15:18
• The derivative of the first term is $0$. For the second, use the quotient rule and the chain rules on each summand . (Assuming $\beta = \beta_p$.) – Ethan Bolker Apr 24 at 15:20
• @martycohen, thanks marty fixed it, – Emily Fassbender Apr 24 at 15:23
• @EthanBolker, thanks Ethan I will try that. – Emily Fassbender Apr 24 at 15:24
• The way you've written this, the answer is zero, because summing something over $t_i\in R_i$ when there's no $t_i$ anywhere in the "something" is the same as multiplying by $\lvert R_i\rvert$. After doing the obvious cancellations you no longer have anything dependent on $\beta.$ I suspect the formula you wrote is not the one you want. – David K Apr 24 at 15:36

I will assume that you meant $$\beta$$ instead of $$\beta_p$$.

If $$f(\beta) = \sum_{i=1}^{n }x_i - \dfrac{\sum_{t_i\in(R_i)}x_ie^{x_i\beta}}{\sum_{t_i\in(R_i)}e^{x_i\beta}}$$, then we just apply standard rules of differentiation.

Writing $$b$$ for $$\beta$$, $$f(b) = \sum_{i=1}^{n }x_i - \dfrac{\sum_{t_i\in(R_i)}x_ie^{x_ib}}{\sum_{t_i\in(R_i)}e^{x_ib}} = \sum_{i=1}^{n }x_i - \dfrac{g(b)}{h(b)}$$, where $$g(b) =\sum_{t_i\in(R_i)}x_ie^{x_ib}$$ and $$h(b) =\sum_{t_i\in(R_i)}e^{x_ib}$$.

Then $$f'(b) =-\dfrac{g'(b)h(b)-g(b)h'(b)}{h^2(b)}$$.

The answer is then given by $$g'(b) =\sum_{t_i\in(R_i)}x_i^2e^{x_ib}$$ and $$h'(b) =\sum_{t_i\in(R_i)}x_ie^{x_ib}$$.

Nothing complicated.

• ,Marty is the numerator of $f'(b)$ = 0 ? – Emily Fassbender Apr 24 at 15:45
• No. It's a double sum because each of g. h. g', and h' are sums. – marty cohen Apr 24 at 15:59