$\displaystyle \displaystyle{ =\frac{d}{dy}\big(\int_0^{\frac{y-b}{a}}\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx\big)=\frac{1}{\sigma\sqrt{2\ pi}}\frac{d}{dy}\big(\int_0^{\frac{y-b}{a}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx\big) \overbrace{=}^{A} \frac{1}{\sigma\sqrt{2\pi}} \frac{d}{dy}\big(h(g(y))\big)}= }$

$\displaystyle \displaystyle{ = \frac{1}{\sigma\sqrt{2\pi}} h'\big(g(y)\big)g'(y) \overbrace{=}^{B} \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(\frac{y-b}{a}-\mu)^2}{2\sigma^2}} \frac{1}{a} = \frac{1}{(a\sigma)\sqrt{2\pi}} e^{-\frac{(y-(a\mu+b))^2}{2(a\sigma)^2}}}$

$\displaystyle \implies Y\sim N(a\mu +b, |a|\sigma)$

Is

__everything__ correct?

**A**: Let $\displaystyle h(y)=\int_0^y e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx$ and $\displaystyle g(y)=\frac{y-b}{a}$.

**B**: $\displaystyle h'(y)=\frac{d}{dy}\big(h(y)\big)=\frac{d}{dy}\big( \int_0^y e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx\big)\overbrace{=}^{FTC} e^{-\frac{(y-\mu)^2}{2\sigma^2}}$.

**FTC ** = **F**undamental **t**heorem of **c**alculus.