Finding the marginal density function of Y Okay, the question is like this: 
$f_{x}(x) = xe^{-x^2/2}$ for all $x>0$ and $Y = \ln X$, find the density of $Y$. 
I don't understand a particular step of this problem. 
First they start for $x \longrightarrow x = e^y$ because the natural log is strictly increasing and is therefore one-to-one. 
Now this is the step that is confusing me: 
$f_y = f_x(v(y))\times|v'(y)|$
where $v(y) = e^y$ 
Can someone help me grasp the intuition of this step? It is really confusing me. Why is this step used to find the mariginal distribution of Y?

An easier way see this is by just looking in terms of CDF's.
$F_Y(y) = P(Y \leq y) = P(\log X \leq y) = P(X \leq e^y) = F_X(e^y)$. 
Then, $f_Y(y) = F_Y'(y) = \frac{d}{dy} [ F_X(e^y) ] = f_X(e^y) \frac{d}{dy} [ e^y ] = e^y f_X(e^y)$. 
That step performs the $F_Y'(y)$ step and accounts for the change of variables you did for writing the CDF of $Y$ (which is an integral of $f_Y(y)$) in terms of the CDF of $X$ (which is also an integral of $f_X(x)$). 
If you have a calculus book handy, look up Jacobians - they will talk about them when doing change of variables for integrals. 
