Calculus, 10th Edition (Anton)

$\frac{6x^2-16x-15}{(3x-4)^2}$
According to the quotient rule, $$\frac{d}{dx}\frac{2x^2+5}{3x-4}=\frac{(3x-4)\frac{d}{dx}(2x^2+5)-(2x^2+5)\frac{d}{dx}(3x-4)}{(3x-4)^2}$$ $$=\frac{(3x-4)(4x)-(2x^2+5)(3)}{(3x-4)^2}$$ Simplified, $$=\frac{(12x^2-16x)-(6x^2+15)}{(3x-4)^2}$$ $$=\frac{6x^2-16x-15}{(3x-4)^2}$$