If $$\Phi(z)=\int_{-\infty}^z\frac 1{2\pi}e^{-t^2/2}dt$$ $$\phi(z)=\frac 1{2\pi}e^{-z^2/2}$$ I need to prove that $\frac d{dx} \Phi(\frac{x-\mu}{\sigma})=\frac 1\sigma \phi(\frac{x-\mu}{\sigma})$.
when I substitute and differentiate I get $\frac d{dx} \Phi(\frac{x-\mu}{\sigma})=\phi(\frac{x-\mu}{\sigma})$. According to my text, there is a chain rule that has to be applied here but I don't know why and how. May you help?
Hint. One may recall that, by applying the standard chain rule, $$ \left[f\left(\frac {x-\mu}\sigma\right)\right]'=\color{red}{\frac 1\sigma}\times f'\left(\frac {x-\mu}\sigma\right). $$