Answer
$y'=\dfrac{10}{(3-4x)^{2}}$
Work Step by Step
$g(x)=\dfrac{1+2x}{3-4x}$
Differentiate using the quotient rule
$g'(x)=\dfrac{(3-4x)(1+2x)'-(1+2x)(3-4x)'}{(3-4x)^{2}}=\dfrac{(3-4x)(2)-(1+2x)(-4)}{(3-4x)^{2}}=...$
Evaluate the products in the numerator
$...=\dfrac{6-8x+4+8x}{(3-4x)^{2}}=\dfrac{10}{(3-4x)^{2}}$
$y'=\dfrac{10}{(3-4x)^{2}}$
