Change of random variables 
Let $X$ be a real-valued random variable with density function
  $$
f_X(x) = \begin{cases}
\frac{4}{15}x^3 & 1\leq x\leq 2 \\
0 & \text{otherwise.}
\end{cases}
$$
Find the density function of $Y =e^X$ and $Z= (X-2)^2$.

I am confused on how to tackle this problem. I know $\mathbb{P}\{Y\le y\} = \mathbb{P}\{e^X\le y\} = \mathbb{P}\{X\leq \ln y\}$.
However then I am lost on what to do. 
 A: We find the cdf $F_Y(y)$ of $Y$, by continuing on the path you were on. For $y$ between $e^1$ and $e^2$, we have
$$F_Y(y)=\Pr(Y\le y)=\Pr(e^X\le y)=\Pr(X\le \ln y)=\int_1^{\ln y} \frac{4}{15}x^3\,dx.\tag{1}$$
For the density function $f_Y(y)$ in the interval $[e^1,e^2]$ we want to differentiate the $F_Y(y)$ of (1). (The density will be $0$ outside this interval.)
Now we have two options: (i) Calculate the integral on the right of (1), and then differentiate or (ii) differentiate directly using the Fundamental Theorem of Calculus and the Chain Rule.
The path (i) is easy, for the integration gives no problem. I suggest also using path (ii), to  prepare for cases where the integration is not easy.
A: Consider the cumulative distribution function of $Y$ and $Z$, and use them (by differentiating) to get the probability density functions. For instance, for $x\in\mathbb{R}$,
$$
\begin{align*}
F_Y(x) &= \mathbb{P}\{Y \leq x\} = \mathbb{P}\{X \leq \ln x\} = \int_{-\infty}^{\ln x} f_X(t)dt \\
&= \begin{cases}
0 & x \leq e\\
1 & x \geq e^2\\
\int_{1}^{\ln x} \frac{4}{15}t^3 dt = \frac{\ln^4 x -1}{15} & \text{o.w.}
\end{cases}
\end{align*}
$$
so that 
$$
f_Y(x) = F^\prime_Y(x) = \begin{cases}
\frac{4}{15}\frac{\ln^3 x}{x}& e\leq x \leq e^2\\
0 & \text{o.w.}
\end{cases}
$$
A: $F_Y(y)=P(Y \leq y) = P(X \leq \ln y) = \int_{-\infty}^{\ln y} f_X(x) dx$. 
For $z <0$, $F_Z(z)=P(Z \leq z) =0$ as $Z \geq 0$. Thus, for $z \geq 0$
$F_Z(z)=P(Z \leq z) = P((X-2)^2 \leq z) = P(|X-2| \leq \sqrt{z}) = P( X-2 \leq \sqrt{z} \text{ and } -(X-2) \leq \sqrt{z}) = P(-\sqrt{z}+2 \leq X \leq  \sqrt{z}+2) = \int_{-\sqrt{z}+2}^{\sqrt{z}+2} f_X(x) dx$.
Now, the density of $Y$ is $f_Y(y) = F_Y'(y)$ and density of $Z$ is $f_Z(z) = F_Z'(z)$. 
