## Calculus: Early Transcendentals 8th Edition

$y'=\dfrac{10}{(3-4x)^{2}}$
$g(x)=\dfrac{1+2x}{3-4x}$ 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 $...=\dfrac{6-8x+4+8x}{(3-4x)^{2}}=\dfrac{10}{(3-4x)^{2}}$ $y'=\dfrac{10}{(3-4x)^{2}}$