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Suppose that I have $f(x) = \text{ln}\Big(1 + \sum_{i=1}^{I}p_i(x)q_i(x)\Big)$. I want to find $f'(x)$. Based on my understanding, I would obtain:

$f'(x) = \frac{1}{1 + \sum_{i=1}^{I}p_i(x)q_i(x)}\frac{d}{dx}\Big(1 + \sum_{i=1}^{I}p_i(x)q_i(x)\Big)$

$f'(x) = \frac{1}{1 + \sum_{i=1}^{I}p_i(x)q_i(x)}\frac{d}{dx}\Big( \sum_{i=1}^{I}p_i(x)q_i(x)\Big)$

However, I do not know how to solve $\frac{d}{dx}\Big( \sum_{i=1}^{I}p_i(x)q_i(x)\Big)$. How to solve it?

bnbfreak
  • 219

1 Answers1

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$$\frac{d}{dx}\Big( \sum_{i=1}^{I}p_i(x)q_i(x)\Big)$$ $$=\frac{d}{dx}\Big( p_1(x)q_1(x)+p_2(x)q_2(x)+\cdots +p_I(x)q_I(x)\Big)$$ $$=p_1'(x)q_1(x)+p_1(x)q_1'(x)+p_2'(x)q_2(x)+p_2(x)q_2'(x)+\cdots+p_I'(x)q_I(x)+p_I(x)q_I'(x)$$ $$=\sum_{i=1}^{I}\Big(p_i'(x)q_i(x)+p_i(x)q_i'(x)\Big)$$

nmasanta
  • 9,222