# $X$~Exp$(\lambda)$ with $Y=cX$ - find the cdf for $Y$

I have this question that I am not sure on how to deal with the $Y=cX$ part, can anyone help?

$X$ is a continuous random variable with $X$~$Exp(\lambda)$ and let $Y=cX$ where $c > 0$ is a fixed constant. I must find the probability density function (pdf) for $Y$ using two different methods:

1) Find the distribution function (cdf) for $Y$ and then differentiate it to find the pdf for $Y$.

I am fine with differentiation, I just am not sure how to find the cdf for $Y$

2) Noting that the function $g(x) = cx$ is strictly increasing and differentiable, find the pdf for $Y$ directly from the pdf for $X$

All I know is that you can write $X$~$\exp(\lambda)$ as:

$$f_X (x)=\left\{ \begin{array}{ll} 0 & \mbox{x \le 0};\\ \lambda e^{(-\lambda x)} & \mbox{x < 0}.\end{array} \right.$$

But I do not know what I can do with this (if anything).

CDF of random variable $Y$ is $P(Y \leq y)$, so we can write $P(Y \leq y) = P(cX \leq y) = P(X \leq \frac{y}{c})$
For the second method you can use $f_Y(y)=\frac{f_X(g^{-1}(y))}{g'(g^{-1}(y))}$