Derivative with a fraction freaks me out

Question

I have this:

$L(\mu, \sigma^2) = -\frac{1}{2\sigma^2}\sum_{n=1}^N(x_n-\mu)^2$

and I need to end up here:

$\frac{\partial}{\partial\sigma^2}L(\mu, \sigma^2) = \frac{1}{2\sigma^4}\sum_{n=1}^N(x_n-\mu)^2$

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But when I try to work things out, I do:

$L(\mu, \sigma^2) = -2^{-1}(\sigma^2)^{-1}\sum_{n=1}^N(x_n-\mu)^2 = -2^{-1}\sum_{n=1}^N(x_n-\mu)^2(\sigma^2)^{-1}$

thus

$\frac{\partial}{\partial\sigma^2}L(\mu, \sigma^2) = \frac{1}{2}\sum_{n=1}^N(x_n-\mu)^2\space\sigma^2$

What am I missing? :/

Answer 1

Think $-2^{-1}\sum_{n=1}^N(x_n-\mu)^2$ is just a multiplier of $\sigma^{-2}$ and notice $\frac{\partial}{\partial\sigma^2} \sigma^{-2} = (-1)(\sigma^2)^{-2}$ .

So you get $\frac{\partial}{\partial\sigma^2}L(\mu, \sigma^2) = -2^{-1}(\sum_{n=1}^N(x_n-\mu)^2) (-1)(\sigma^2)^{-2} = \frac{1}{2\sigma^4}\sum_{n=1}^N(x_n-\mu)^2 $.