Find $f_W$ for $W=-\frac{1}{2}\log(X)$. Given answer doesn't match my answer 
Let $X$ a random variable with the p.d.f.
$$ f_X(x) = \begin{cases}
3x^2, & \text{if } 0<x<1 \\
0, & \text{otherwise}
\end{cases}$$
and $W=-\frac{1}{2}\log(X) $. Find the p.d.f. $f_W$ for $W$.

I've been given this task and the answer to the question as well. So, my idea was to use the formula:
$$f_Y(y)=\frac{f_X(x_1)}{g'(x_1)}$$
where $\ g(x_1)=y$.
However, the answer that I am ending up with simply does not align with the answer given by the instructor. I am told that the answer should be $W \sim \operatorname{Exp}(6) $.
This is how I'm breaking the task down:
$$ f_X(x) = \begin{cases}
3x^2, & \text{if } 0<x<1 \\
0, & \text{otherwise}
\end{cases}
\qquad
\text{and}
\qquad
g(x) = -\frac{1}{2}\log(x)$$
are given by the task. So what I need to do is differentiate $g(x)$ and find the inverse function of $W$ and put it all together:


*

*Derivative: $g'(x) = -\frac{1}{2x} $.

*Inverse: $x_1 = g^{-1}(x) = e^{-2x}$.


Putting it together:
$$ f_W(x)
= \frac{3 \cdot (e^{-2x})^2}{-1/(2 \cdot e^{-2x})}
=-6(e^{-2x})^3 $$
As you can see this doesn't match the given answer. I've used hours trying other methods but cannot find the mistake. Any thoughts?
Best regards
 A: The cdf of  $W$ can be calsulated as follows
$$F_W(w)=P(W<w)=P\left(-\frac12\log(X)<w\right)=$$
$$=P\left(\log(X)>-2w\right)=P\left(X>e^{-2w}\right)=$$
$$=3\int_{e^{-2w}}^1x^2\ dx=\begin{cases}0&\text{ if } &w<0\\1-e^{-6w}&\text{ if } &0\leq w.\end{cases}$$
The density is then
$$f_W(w)=\frac{d F_W}{dw}=6e^{-6w}$$
if $w\geq 0$.
A: The pdf of $W$ can be written as
$$
f_W(w)=\int_0^1 dx\ 3x^2\delta\left(w+\frac{1}{2}\ln x\right)\ ,
$$
using a Dirac delta. Since $w+(1/2)\ln x=0\Rightarrow x=e^{-2w}$, we have
$$
f_W(w)=\int_0^1 dx\ 3x^2\frac{\delta\left(x-e^{-2w}\right)}{\frac{1}{2e^{-2w}}}=6e^{-2w}\times e^{-4w}\mathbb{1}(e^{-2w}<1)=6e^{-6 w}\theta(w)\ ,
$$
which is correctly normalized over $w\in (0,\infty)$. The fact that the support is $(0,\infty)$ is evident from the definition, as the $\log$ of a number $X$ between $0$ and $1$ is negative. I think all you are missing is the absolute value of $g'(x)$.
PS I highly doubt that your instructor gave you $w\sim e(6)$ as a solution. That notation doesn't make any sense, to start with...
A: Let $w\in (0,\infty)$. Let $F_W$ be the CDF of of $W$ and $F_X$ the CDF of $X$
\begin{align}
F_W(w)&=P(W<w)=P\left(\frac{-\ln(X)}{2}<w\right)\\
&=P(\ln(X)>-2w)=P(X>e^{-2w})\\&
=1-P(X\leq e^{-2w})=1-F_X(e^{-2w})
\end{align}
Differentiating and using the Fundamental Theorem of Calculus gives us the pdf of $W$. So:
\begin{align}
f_W(w)=2e^{-2w}f_X(e^{-2w})=2e^{-2w}3e^{-4w}=6e^{-6w}
\end{align}
since $w\in(0,\infty)$. It can be easily seen that $f_W(w)=0$ for $w\leq 0$. 
