How to find $\partial\chi^2/\partial a$ where $\chi^2=\sum_{i=1}^N\frac{D(x_i)-a-b(x_i)^2}{\sigma_i^2}$?

Question

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} &= \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}\\ &=\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}\\ &=\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\\ &=\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\\ &=\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} &= \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]\\ &=\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]\\ &= 0-\dfrac{\partial}{\partial a}\Bigg[\dfrac{-a}{\sigma_1^2}+\dfrac{-a}{\sigma_2^2}+\cdots+\dfrac{-a}{\sigma_N^2}\Bigg]\\ &= -\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]\\ &= -\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]\\ &= -\Bigg[\dfrac{-1}{\sigma_1^2}+\dfrac{-1}{\sigma_2^2}+\cdots+\dfrac{-1}{\sigma_N^2}\Bigg]\\ &= \sum_{i=1}^{N}\dfrac{1}{\sigma_i^2}\\ \end{align}

Answer 1

Nope. If $D$, $b$ and $\sigma$ are independet of $a$, then the partial differential of their combination should be zero. Second mistake is the sign you used. (between sum signs it should be "+"). However later you corrected it. Next you can't take our $\frac{1}{\sigma_{i}^{2}}$ out of the sum sign (as $\sigma_i$-s depends on $i$). And the last mistake is that sum of $a$-s is not $a^{N}$, but rather $aN$. So as to me the answer should be $\sum_{i=1}^{N}\frac{1}{\sigma_{i}^{2}}$ (because partial of $\frac{a}{\sigma_{i}^{2}}$ is $\frac{1}{\sigma_{i}^{2}}$).

If $D$, $b$ and $\sigma$ depends on $a$(As it may when the initial problem comes from statistics), then the dependence form is crucial. (Without it you can't explicitly tell what the partial will be.)

Sorry(too long to type it as comment).