#### Answer

$y'=\dfrac{c}{(1+cx)^2}$

#### Work Step by Step

$y=\dfrac{cx}{1+cx}$
$c$ is a constant and $x$ is a variable.
Use the quotient rule:
$y'=\dfrac{(1+cx)(c)-(cx)(0+c)}{(1+cx)^2}$
The derivative of $cx$ is $c$ because $c$ is a constant so we keep it and the derivative of $x$ is $1$ since $x$ is a variable. $c\bullet 1=c$ This can also be proven using the product rule.
Distribute the numerator:
$y'=\dfrac{c+c^2x-c^2x}{(1+cx)^2}$
Combine like terms:
$y'=\dfrac{c}{(1+cx)^2}$