differentiating the numerator
Quantitative finance Qn 2 ZA 2014
you know that the PDF of the standard normal function is
\( \phi(z)=\frac{1}{\sqrt{2\pi}}e^{\frac{-z^{2}}{2}} \)
So just let z=y and substitute this into the numerator part,
\(\frac{\phi(y)}{y} = \frac{1}{y}\frac{1}{\sqrt{2\pi}}e^{\frac{-y^{2}}{2}}\)
differentiate the numerator using product rule:
\(\frac{\mathrm{d} }{\mathrm{d} y}(\frac{1}{y}\frac{1}{\sqrt{2\pi}}e^{\frac{-y^{2}}{2}})= \frac{1}{y}(\frac{1}{\sqrt{2\pi}}e^{\frac{-y^{2}}{2}})(-2y)(\frac{1}{2})+(\frac{1}{\sqrt{2\pi}}e^{\frac{-y^{2}}{2}})(\frac{-1}{y^{2}})\)
and after simplifying
\((\frac{1}{\sqrt{2\pi}}e^{\frac{-y^{2}}{2}})(-1-\frac{1}{y^{2}})=-(1+\frac{1}{y^{2}})(\frac{1}{\sqrt{2\pi}}e^{\frac{-y^{2}}{2}})\)
since \(\phi(y) = \frac{1}{\sqrt{2\pi}}e^{\frac{-y^{2}}{2}}\)
so \((\frac{1}{\sqrt{2\pi}}e^{\frac{-y^{2}}{2}})(-1-\frac{1}{y^{2}})=-(1+\frac{1}{y^{2}})\phi(y)\)
and that's the differentiation working for the numerator part.
hope there's no typo...