## Calculus 8th Edition

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