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I was doing the problem provided in the picture but I do not understand how do they obtain the answer. I am not sure how to differentiate the sum. I end up getting: alpha - 1 - 1/K. I believe I need to use the chain rule but I am not quite sure about implicit derivatives.

Your input is very much appreciated.

The question

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There is no problem around implicit differentiation.

Your formula is $$p_i=\frac{Aq_i^{\alpha-1}}K$$ Take logarithms of both sides to get $$\log(p_i)=\log(A)+(\alpha-1)\log(q_i)-\log(K)$$ Differentiate with respect to $q_i$ and get $$\frac 1{p_i}\frac{dp_i}{dq_i}=\frac{\alpha-1}{q_i}-\frac 1K\frac{dK}{dq_i}$$ Now, consider $$K=\sum_{j=1}^n q_j^\alpha$$ So $$\frac{dK}{dq_i}=\alpha q_i^{\alpha-1}$$ I am sure that you can take from here.

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  • $\begingroup$ Makes it clearer now. Thanks. In general, we do not need any special adjustments if we deal with a derivative of summation - dK/dqi? $\endgroup$ – Lawrence May 23 '15 at 12:33
  • $\begingroup$ The derivative of a sum is the sum of the derivatives, isn't it ? When you are not sure, just write two or three terms. I am sure it will become clearer. Cheers :-) $\endgroup$ – Claude Leibovici May 23 '15 at 12:36
  • $\begingroup$ Then if we sum to 'n' why we do not have n constant in front of the expression and how 'j' transforms into 'i'? I know these questions sound very stupid but I have a gap in derivatives since the school days and I have never managed to fill it in... I really appreciate your kind help. $\endgroup$ – Lawrence May 23 '15 at 13:04
  • $\begingroup$ You are very welcome ! When you differentiate with respect to $q_i$, consider that all other $q_j$ $(j\neq i)$ terms are constants. Is this better ? If not, just post. $\endgroup$ – Claude Leibovici May 23 '15 at 13:07
  • $\begingroup$ Oh yes! now it makes sense! Thank you so much! $\endgroup$ – Lawrence May 23 '15 at 13:09

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