Derivative of capital Pi product I wanted to find the derivative of this function at $x=6$
$$y= \prod_{i=1}^{10} (x-i) = (x-1)(x-2) \cdots (x-10) $$
without expanding all of the brackets, so I used the product rule to find a pattern. However, the resulting sum tells me that the derivative is zero at every whole number which is obviously not true. I've been over my solution and I can't see how I've gone wrong. Please could someone highlight where I went wrong? Thank you in advance.
\begin{align*}
      \frac{\textit{d}y}{dx} &= (x-2)(x-3) \cdots (x-10) + (x-1) \frac{d}{dx} \biggl((x-2) \cdots (x-10) \biggr) \\
       &= \prod_{i=2}^{10} (x-i) +  (x-1) \frac{d}{dx} \biggl(\prod_{i=2}^{10} (x-i) \biggr) \\
       &= \prod_{i=2}^{10} (x-i) +  (x-1)\prod_{i=3}^{10} (x-i) + (x-1)(x-2)\frac{d}{dx} \biggl(\prod_{i=3}^{10} (x-i) \biggr) \\
                &= \prod_{i=2}^{10} (x-i) +  (x-1)\prod_{i=3}^{10} (x-i) + (x-1)(x-2)\biggl(\prod_{i=4}^{10} (x-i) \biggr) + \cdots \\
                      &= \frac{y}{x-1} + \frac{y}{x-2} + \frac{y}{x-3}+\cdots + \frac{y}{x-10} \\
                             &= \sum_{i=1}^{10} \biggl(\frac{y}{x-i}\biggr)
    \end{align*}

The pattern is simpler to understand than this. In general suppose $f(x) = (x - a) g(x)$ and we want to compute $f'(a)$. (Here $g(x)$ could be a polynomial or more generally any differentiable function.) Then
$$f'(x) = g(x) + (x - a) g'(x)$$
which gives $\boxed{ f'(a) = g(a) }$. This is particularly easy if you notice that applying the definition of the derivative in this case directly gives
$$f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} = \lim_{x \to a} g(x) = g(a).$$
