Finding Density From Expected Value Problem:
Given that $X \sim N(0,1)$, let $Y = e^{X}$. Find a formula for the density, $f_{Y}(y)$.
Progress:
Using the Law of the Unconscious Statistician, I computed
$$E[Y] = E[e^{X}] = \sqrt{e}.$$ 
I figured that a potential solution had to satisfy both:
$$\int_{\mathrm{R}} f_{Y}(y) \, dy = 1$$
$$\int_{\mathrm{R}} yf_{Y}(y) \, dy = \sqrt{e}$$
I tried playing around with integration by parts on the second one but realized that wasn't going to get me anywhere because I can't find $f'$ or $\int f \, dy$.

Now that I'm typing it out, maybe I can make integration by parts work using
$$F(y) = \int_{-\infty}^{t} f_{Y}(t) \, dt$$
If anyone has some insight that would be really helpful. I'm probably on the wrong track but I'm gonna go play around with this some more.
 A: $F_{Y}(y) = \mathbb{P}\left(e^{X} \leq y\right) = \mathbb{P}\left(X \leq \ln y\right) = F_{X}(\ln y) = \Phi\circ\ln (y)$.
By chain rule, $f_{Y}(y) = \frac{\mathrm{d}}{\mathrm{d}y}\left(\Phi\circ\ln (y)\right) = \Phi^{\prime}\circ\ln(y)\:\frac{\mathrm{d}}{\mathrm{d}y}\ln y = \frac{\phi\circ\ln(y)}{y} = \frac{1}{\sqrt{2\pi}} \frac{e^{-\frac{1}{2}\left(\ln y\right)^{2}}}{y}$.
A: For any differentiable and monotonic (hence invertible) function $Y=g(X)$ we have the relationship
$$f_Y(y)=\frac{f_X(x)}{|g'(x)|}=\frac{f_X(g^{-1}(y))}{|g'(g^{-1}(y))|}$$
In our case, $g^{-1}(y) = \log Y$. Hence
$$ f_Y(y) = \frac{e^{-x^2/2}}{\sqrt{2 \pi}}\frac{1}{e^x}=\frac{e^{-(\log y)^2/2}}{y \sqrt{2 \pi}} = \frac{y^{-\log(y)/2-1}}{\sqrt{2 \pi}} \hspace{1cm} y> 0$$
A: \begin{align}
f_Y(y) & = \frac d {dy} \Pr(Y\le y) = \frac d {dy} \Pr(e^X \le y) = \frac d {dy} \Pr(X \le \log y) = \frac d {dy} F_X(\log y) \\[10pt]
& \underbrace{{} = f_X(\log y) \cdot \frac d {dy} \log y}_\text{chain rule} {} = \frac 1 {\sqrt{2\pi}} e^{-(\log y)^2/2} \cdot \frac 1 y.
\end{align}
