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} &amp;= \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]\\     &amp;=  \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]\\     &amp;= 0-\dfrac{\partial}{\partial b}\Bigg[\sum_{i=1}^N\dfrac{b(x_i)^2}{\sigma_i^2}\Bigg]\\     &amp;= -\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]\\     &amp;= -\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]\\     &amp;= -\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]\\     &amp;= -\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)\\     &amp;= -\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]\\     &amp;= -\sum_{i=1}^N\dfrac{2b(x_i)}{\sigma_1^2} \end{align*}  Is this correct?  EDIT: I previously asked  question. But the fact that $b$ is a function and changes with each different $x_i$ confuses me.
How to find $\partial\chi^2/\partial a$ where $\chi^2=\sum_{i=1}^N\dfrac{D(x_i)-a-b(x_i)^2}{\sigma_i^2}$?  My attempt: \begin{align} \dfrac{\partial}{\partial a}\sum_{i=1}^N\dfrac{D(x_i)-a-b(x_i)^2}{\sigma_i^2} &amp;= \dfrac{\partial}{\partial a}\sum_{i=1}^N\dfrac{D(x_i)-b(x_i)^2}{\sigma_i^2}-\sum_{i=1}^N\dfrac{-a}{\sigma_i^2}\\ &amp;=\sum_{i=1}^N\dfrac{D(x_i)-b(x_i)^2}{\sigma_i^2}-\dfrac{\partial}{\partial a}\sum_{i=1}^N\dfrac{-a}{\sigma_i^2}\\ &amp;=\sum_{i=1}^N\dfrac{D(x_i)-b(x_i)^2}{\sigma_i^2}-\dfrac{1}{\sigma_i^2}\cdot\dfrac{\partial}{\partial a}\sum_{i=1}^N-a\\ &amp;=\sum_{i=1}^N\dfrac{D(x_i)-b(x_i)^2}{\sigma_i^2}-\dfrac{1}{\sigma_i^2}\cdot\dfrac{\partial}{\partial a}a^N\\ &amp;=\sum_{i=1}^N\dfrac{D(x_i)-b(x_i)^2}{\sigma_i^2}-\dfrac{1}{\sigma_i^2}\cdot N\cdot a^{N-1}\\ \end{align}  Is this correct?  EDIT: As per the accepted answer, this is how I got my final partial derivative. \begin{align}             \dfrac{\partial}{\partial a}\sum_{i=1}^N\dfrac{D(x_i)-a-b(x_i)^2}{\sigma_i^2} &amp;= \dfrac{\partial}{\partial a}\Bigg[\sum_{i=1}^N\dfrac{D(x_i)-b(x_i)^2}{\sigma_i^2}-\sum_{i=1}^N\dfrac{-a}{\sigma_i^2}\Bigg]\\             &amp;=\dfrac{\partial}{\partial a}\Bigg[\sum_{i=1}^N\dfrac{D(x_i)-b(x_i)^2}{\sigma_i^2}\Bigg]-\dfrac{\partial}{\partial a}\Bigg[\sum_{i=1}^N\dfrac{-a}{\sigma_i^2}\Bigg]\\             &amp;= 0-\dfrac{\partial}{\partial a}\Bigg[\dfrac{-a}{\sigma_1^2}+\dfrac{-a}{\sigma_2^2}+\cdots+\dfrac{-a}{\sigma_N^2}\Bigg]\\             &amp;= -\Bigg[\dfrac{\partial}{\partial a}\Bigg[\dfrac{-a}{\sigma_1^2}\Bigg]+\dfrac{\partial}{\partial a}\Bigg[\dfrac{-a}{\sigma_2^2}\Bigg]+\cdots+\dfrac{\partial}{\partial a}\Bigg[\dfrac{-a}{\sigma_N^2}\Bigg]\Bigg]\\             &amp;= -\Bigg[\dfrac{-1}{\sigma_1^2}\cdot\dfrac{\partial}{\partial a}[a]+\dfrac{-1}{\sigma_1^2}\cdot\dfrac{\partial}{\partial a}[a]+\cdots+\dfrac{-1}{\sigma_N^2}\cdot\dfrac{\partial}{\partial a}[a]\Bigg]\\             &amp;= -\Bigg[\dfrac{-1}{\sigma_1^2}+\dfrac{-1}{\sigma_2^2}+\cdots+\dfrac{-1}{\sigma_N^2}\Bigg]\\             &amp;= \sum_{i=1}^{N}\dfrac{1}{\sigma_i^2}\\         \end{align}
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