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I am having trouble understanding the following derivation. enter image description here

Specifically, I am not able to understand why the partial derivative of $x_i$ with respect to $s_k$ is what it is. My hunch is that it is has something to do with the rule $ log(\frac{a}{b}) = log(a) - log(b)$. The reason being that if I am subtracting a squared value(which I generally would have expected to seen in the denominator), then it probably has something to do with logarithms. However, I am not able to derive the answer in any way.

Jonathan
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1 Answers1

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If you apply the rule of derivation for a quotient : $ \left( \frac{u}{v} \right)' = \frac{u'v-uv'}{v^2} $, on $x_i$ it gives :

$$ \frac{\partial x_i}{\partial s_k}=\frac{\delta_{ik}e^{s_k}\left(\sum_c e^{s_c}\right)-e^{s_i}e^{s_k}}{\left(\sum_c e^{s_c}\right)^2} $$ where $\delta_{ik}=1$ if $i=k$ and $0$ if $i\neq k$.

This is exactly what you want.

L.Mathis
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  • Sorry, I'm not understanding why there is a summation in the numerator? – Jonathan Oct 06 '16 at 19:47
  • Here you apply the formula with $u=e^{s_i}$ and $v=\sum_c e^{s_c}$ so the sum on the numerator is the $v$ of $u'v$. – L.Mathis Oct 06 '16 at 19:55
  • Also, the $\delta_{ik}$ means that if you derive by $s_k$ all the $s_i$ will be constant except $s_k$ – L.Mathis Oct 06 '16 at 20:00
  • In which case, the summation doesn't appear in the numerator of the answer given. Am I just missing something? – Jonathan Oct 06 '16 at 20:01
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    Ah i misunderstood you sorry. you separate in a sum of two ratio and cancel the sums. For example if $i=k$ it gives : $ \frac {e^{s_i} \left(\sum e^{s_c}\right)}{\left(\sum e^{s_c}\right)^2}-\frac{(e^{s_i})^2}{\left(\sum e^{s_c}\right)^2}=\frac {e^{s_i} }{\left(\sum e^{s_c}\right)}-\frac{(e^{s_i})^2}{\left(\sum e^{s_c}\right)^2}$ – L.Mathis Oct 06 '16 at 20:02
  • That makes so much sense! Thank you! – Jonathan Oct 06 '16 at 20:06