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. :-)