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Differentiating

$\int_{(m-d-\mu)/\sigma}^{\infty}xf(x)dx$

with respect to $\sigma$, where $\sigma$ is the standard deviation of the standardarized random variable $x$ and $\mu$ its mean. I guess that it is

$\frac{m-d-\mu}{\sigma}f\left(\frac{m-d-\mu}{\sigma}\right).$

Correct me if I am wrong.

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    Ah, sorry, I wrote my comment before I saw your response to yohBS. But in the end you are missing also a factor of $m-d-\mu$ in addition to the $\sigma^{-2}$.2012-09-27

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Let $F(y) = \int_y^\infty x f(x)~ \mathrm{d}x$, we can write it also as

$ F(y) = - \int_{\infty}^y x f(x)~ \mathrm{d}x$

so by the fundamental theorem of calculus

$ \frac{\mathrm{d}}{\mathrm{d}y} F(y) = - y f(y) $

Let $G(\sigma) = F( (m-d-\mu)/\sigma )$. Then we have that by the Chain rule (not the Leibniz rule!) that

$ \frac{\mathrm{d}}{\mathrm{d}\sigma} G(\sigma) = \frac{\mathrm{d}}{\mathrm{d}\sigma} F\left( \frac{m - d - \mu}{\sigma}\right) = F'\left( \frac{m - d - \mu}{\sigma}\right) \cdot \frac{\mathrm{d}}{\mathrm{d}\sigma} \frac{m-d-\mu}{\sigma} $

and you can take it from here. :-)

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    That looks okay to me, provided all the parentheses are correctly closed (I didn't check that).2012-09-27