# Thread: Differentiation of Summation operators

1. ## Differentiation of Summation operators

Hello all,

I am having a hard time understanding how to differentiate summation operators. More specifically, I was given the following solution that I do not understand. Any step by step advice (or a link) would be fantastic....

$\displaystyle$\dfrac{d}{dx}$$\sum\nolimits_{i}p_{i}x_{i}=p_{i}$$ I do not understand why this is equal to p, not the sum of p

Thank you kindly for the help,

Nick

2. May be that the formula is...

$\displaystyle \displaystyle \frac{d}{d x_{i}} \sum_{i} p_{i}\ x_{i} = p_{i}$ (1)

... and if the $\displaystyle x_{i}$ are independent variables that's true...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. What if $\displaystyle \dfrac{d}{dx_i}$$\sum\nolimits_{i}p_{i}x_{i}=p_{i} ? 4. Originally Posted by salohcin Hello all, I am having a hard time understanding how to differentiate summation operators. More specifically, I was given the following solution that I do not understand. Any step by step advice (or a link) would be fantastic.... \displaystyle \dfrac{d}{dx}$$\sum\nolimits_{i}p_{i}x_{i}=p_{i}$$As the others are implying, what you have written makes no sense. The sum depends on many different variables, labeled$\displaystyle x_1$,$\displaystyle x_2$, ...,$\displaystyle x_n$but you are differentiating with respect to "x" which is not any of the variables! What you have is something like$\displaystyle \frac{d(p_1y+ p_2z)}{dx}$where you are differentiating a function of y and z with respect to "x". That derivative is, in fact, 0! But the partial derivatives of$\displaystyle \frac{\partial (p_1x_1+ p_2x_2+ p_3x_3)}{\partial x_1}= p_1$,$\displaystyle \frac{\partial (p_1x_1+ p_2x_2+ p_3x_3)}{\partial x_2}= p_2$, and$\displaystyle \frac{\partial (p_1x_1+ p_2x_2+ p_3x_3)}{\partial x_3}= p_3\$

I do not understand why this is equal to p, not the sum of p

Thank you kindly for the help,

Nick