Why do I need the derivative in variable substitution for probability density function? If $y = g(x)$ and $f_X(x)$ the probability density function of X
$x = g^{-1}(y);$
Then 
$f_Y(y) = f_X(g^{-1}(y)) * (g^{-1})'(y)$
Why do I need to multiply by $(g^{-1})'(y)$?
And not simply:
$f_Y(y) = f_X(g^{-1}(y))$
 A: Following the link from your comment: 
Let $X$ be a continuous random variable with density $f$ such that $f(x)>0$ for $c_1<x<c_2$. Let $u:(c_1,c_2)\to\mathbb{R}$ be a continuous, increasing function and let $v=u^{-1}$ denote its inverse. Define a new random variable $Y=u(X)$ (which indeed is a random variable, because $u$ is measurable), then the cumulative distribution function of $Y$ is given by
$$
F_Y(y)=P(Y\leq y)=P(u(X)\leq y)=P(X\leq v(y))=F_X(v(y)),
$$
when $u(c_1)<y<u(c_2)$. Here we have in the third equality used that $u$ is indeed increasing. To obtain the density of $Y$ we must differentiate $F_Y$ once, and because $F_Y=F_X\circ v$ is a composite function, we can apply the Chain Rule:
$$
f_Y(y)=F_Y'(y)=F_X'(v(y))\cdot v'(y)=f_X(v(y))\cdot v'(y),\quad u(c_1)<y<u(c_2),
$$
and since $v'(y)=(u^{-1})'(y)=\frac{1}{u'(v(y))}$ we can actually write it as
$$
f_Y(y)=f_X(v(y))\frac{1}{u'(v(y))},\quad u(c_1)<y<u(c_2).
$$
Note that we haven't assumed that $u$ is differentiable, but this is implied by the continuity and monotonicity.
A: Informal outline: 

If $Y=g(X)$  the events "X falls in the red interval" and "Y falls in the blue interval" are equivalent, and hence they have same probability. But those probabilities are approximately $f_X(x) Dx$, $f_Y(y) Dy$, and then, assuming $g(X)$ is smooth and monotone, and the intervals are small,
$$f_Y(y)= \frac{f_X(x)}{Dy/Dx}\approx \frac{f_X(x)}{|g'(x)|}=\frac{f_X(g^{-1}(y))}{|g'(g^{-1}(y))|}$$
A: The random variable $X$ has probability density $x\mapsto f_X(x)$ if for all $a<b$ the relation
$$P[a<X<b]=\int_a^b f_X(x)\ dx$$
holds.
Given $X$ and a monotone increasing function $g:\ {\mathbb R}\to{\mathbb R}$ one can consider the new random variable $Y:=g(X)$. It's density $y\mapsto f_Y(y)$ is characterized by the condition
$$P[\alpha<Y<\beta]=\int_\alpha^\beta f_Y(y)\ dy\ .\qquad(1)$$
Now $\alpha<Y<\beta$ is equivalent with $g^{-1}(\alpha)<X<g^{-1}(\beta)$. Therefore we also have
$$\eqalign{P[\alpha<Y<\beta] &=P[g^{-1}(\alpha)<X<g^{-1}(\beta)]=\int_{g^{-1}(\alpha)}^{g^{-1}(\beta)} f_X(x)\ dx\cr &=\int_\alpha^\beta f_X\bigl(g^{-1}(y)\bigr) \bigl(g^{-1}\bigr)'(y)\ dy\ .\cr}\qquad(2)$$
As $(1)$ and $(2)$ hold for arbitrary $\alpha<\beta$ it follows that necessarily
$$f_Y(y)\ \equiv\ f_X\bigl(g^{-1}(y)\bigr) \bigl(g^{-1}\bigr)'(y)\ .$$
