How do I find How to find $\partial\chi^2/\partial b$ when $\chi^2=\sum_{i=1}^N\dfrac{D(x_i)-a-b(x_i)^2}{\sigma_i^2} $?
My attempt: \begin{align*} \dfrac{\partial}{\partial b}\sum_{i=1}^N\dfrac{D(x_i)-a-b(x_i)^2}{\sigma_i^2} &= \dfrac{\partial}{\partial b}\Bigg[\sum_{i=1}^N\dfrac{D(x_i)-a}{\sigma_i^2}-\sum_{i=1}^N\dfrac{b(x_i)^2}{\sigma_i^2}\Bigg]\\ &= \dfrac{\partial}{\partial b}\Bigg[\sum_{i=1}^N\dfrac{D(x_i)-a}{\sigma_i^2}\Bigg]-\dfrac{\partial}{\partial b}\Bigg[\sum_{i=1}^N\dfrac{b(x_i)^2}{\sigma_i^2}\Bigg]\\ &= 0-\dfrac{\partial}{\partial b}\Bigg[\sum_{i=1}^N\dfrac{b(x_i)^2}{\sigma_i^2}\Bigg]\\ &= -\dfrac{\partial}{\partial b}\Bigg[\dfrac{b(x_1)^2}{\sigma_1^2}+\dfrac{b(x_2)^2}{\sigma_2^2}+\cdots+\dfrac{b(x_N)^2}{\sigma_N^2}\Bigg]\\ &= -\dfrac{\partial}{\partial b}\Bigg[\dfrac{b(x_1)^2}{\sigma_1^2}\Bigg]+\dfrac{\partial}{\partial b}\Bigg[\dfrac{b(x_2)^2}{\sigma_2^2}\Bigg]+\cdots+\dfrac{\partial}{\partial b}\Bigg[\dfrac{b(x_N)^2}{\sigma_N^2}\Bigg]\\ &= -\dfrac{1}{\sigma_1^2}\dfrac{\partial}{\partial b}\Bigg[b(x_1)^2\Bigg]-\dfrac{1}{\sigma_2^2}\dfrac{\partial}{\partial b}\Bigg[b(x_2)^2\Bigg]-\cdots-\dfrac{1}{\sigma_N^2}\dfrac{\partial}{\partial b}\Bigg[b(x_N)^2\Bigg]\\ &= -\dfrac{1}{\sigma_1^2}\cdot2\cdot b(x_1)-\dfrac{1}{\sigma_2^2}\cdot2\cdot b(x_2)-\cdots-\dfrac{1}{\sigma_N^2}\cdot2\cdot b(x_N)\\ &= -\Bigg[\dfrac{2}{\sigma_1^2}\cdot b(x_1)+\dfrac{2}{\sigma_2^2}\cdot b(x_2)+\cdots+\dfrac{2}{\sigma_N^2}\cdot b(x_N)\Bigg]\\ &= -\sum_{i=1}^N\dfrac{2b(x_i)}{\sigma_1^2} \end{align*}
Is this correct?
EDIT: I previously asked this question. But the fact that $b$ is a function and changes with each different $x_i$ confuses me.
I figured out the answer thanks to @MPW 's comment.
\begin{align*} \dfrac{\partial}{\partial b}\sum_{i=1}^N\dfrac{D(x_i)-a-b(x_i)^2}{\sigma_i^2} &= \dfrac{\partial}{\partial b}\Bigg[\sum_{i=1}^N\dfrac{D(x_i)-a}{\sigma_i^2}-\sum_{i=1}^N\dfrac{b(x_i)^2}{\sigma_i^2}\Bigg]\\ &= \dfrac{\partial}{\partial b}\Bigg[\sum_{i=1}^N\dfrac{D(x_i)-a}{\sigma_i^2}\Bigg]-\dfrac{\partial}{\partial b}\Bigg[\sum_{i=1}^N\dfrac{b(x_i)^2}{\sigma_i^2}\Bigg]\\ &= 0-\dfrac{\partial}{\partial b}\Bigg[\sum_{i=1}^N\dfrac{b(x_i)^2}{\sigma_i^2}\Bigg]\\ &= -\dfrac{\partial}{\partial b}\Bigg[\dfrac{b(x_1)^2}{\sigma_1^2}+\dfrac{b(x_2)^2}{\sigma_2^2}+\cdots+\dfrac{b(x_N)^2}{\sigma_N^2}\Bigg]\\ &= -\dfrac{\partial}{\partial b}\Bigg[\dfrac{b(x_1)^2}{\sigma_1^2}\Bigg]-\dfrac{\partial}{\partial b}\Bigg[\dfrac{b(x_2)^2}{\sigma_2^2}\Bigg]-\cdots-\dfrac{\partial}{\partial b}\Bigg[\dfrac{b(x_N)^2}{\sigma_N^2}\Bigg]\\ &= -\dfrac{1}{\sigma_1^2}\dfrac{\partial}{\partial b}\Bigg[b(x_1)^2\Bigg]-\dfrac{1}{\sigma_2^2}\dfrac{\partial}{\partial b}\Bigg[b(x_2)^2\Bigg]-\cdots-\dfrac{1}{\sigma_N^2}\dfrac{\partial}{\partial b}\Bigg[b(x_N)^2\Bigg]\\ &= -\dfrac{(x_1)^2}{\sigma_1^2}\dfrac{\partial}{\partial b}[b]-\dfrac{(x_2)^2}{\sigma_2^2}\dfrac{\partial}{\partial b}[b]-\cdots-\dfrac{(x_N)^2}{\sigma_N^2}\dfrac{\partial}{\partial b}[b]\\ &= -\dfrac{(x_1)^2}{\sigma_1^2}\cdot1-\dfrac{(x_2)^2}{\sigma_2^2}\cdot1-\cdots-\dfrac{(x_N)^2}{\sigma_N^2}\cdot1\\ &= -\Bigg[\dfrac{(x_1)^2}{\sigma_1^2}+\dfrac{(x_2)^2}{\sigma_2^2}+\cdots+\dfrac{(x_N)^2}{\sigma_N^2}\Bigg]\\ &= -\sum_{i=1}^N\dfrac{(x_i)^2}{\sigma_i^2} \end{align*}