mapping random variables Let $x$ be a random variable (RV), $a<x<b$ with a pdf $f(x)$. Let's construct a function on $x$, $y=h(x)$ which is continuous and differentiable. 
If an inverse function exists for $h$, say $g(y)=x$. Then it says,
$$f(h(x)) = f(g(y)) \left| \frac{dg(y)}{dy} \right|$$
I understand the mapping through $g(y)$. But how does the $\left| \frac{dg(y)}{dy} \right|$ come into play here? Why do I need the derivative? Perhaps a proof would help but is there a geometric intuition?
 A: One way to understand it is that if $y=h(x)$ is a strictly increasing continuous and differentiable function then $g(y)=h^{-1}(y)$ is too
then $\displaystyle \Pr(Y \le y)=\Pr(X \le g(y))=\int_a^{g(y)} f(x)\,dx$
and if you now take the derivative of that with respect to $y$, you get $f(g(y))\dfrac{dg(y)}{dy}$ 
though your original question is marginally more complicated than this.
A: This method is easiest understand if $h$ is also monotone increasing on $(a,b)$.
Consider $X \sim Unif(0,1)$ so the $f_X(x) = 1$, for $0 <x<1,$ and 0 otherwise.
Also, let $Y = h(X) = X^2.$ Then 
$$P\{0 < X < .1\} = P\{0 < Y < .01\} = 1/10$$
and
$$P\{.9 < X < 1\} = P\{.81 < Y < 1\} = 1/10.$$
Notice that the intervals for $X$ are of equal width, while the intervals
for $Y$ are of very different widths. So it is already clear that the density function $f_Y$ has to be much "taller" above $(0, .01)$ than it is above $(.80,1)$ for these probabilities to hold.
It turns out that $f_Y(y) = 0.5y^{-0.5},$ for $0 < y < 1,$ and 0 otherwise.
This density function $f_Y$ goes asymptotically to infinity near 0, and is about
1/2 near 1.
I have chosen this example because, in this case, the original density function is a constant over its support $(0, 1)$, the entire form of the density of $f_Y$ is due to $dh^{-1}/dy = dx/dy.$ 
(Because the transformation is monotonically increasing, no absolute
value signs are needed.) Intuitively, the derivative can be considered
as answering the question, "What adjustment needs to be made in going from $f_X$ above interval $\Delta x$ so that $f_Y$ accommodates to the change
in the length $\Delta y$ to keep the amount of probability the same for the two tiny intervals?"
Note: In this example $X \sim Beta(1,1) = Unif(0,1)$ and $Y \sim Beta(.5,1).$
