## Calculus with Applications (10th Edition)

$\frac{-30x^4+4x-132x^3+8}{(3x^3+2)^5}$
We are given $y=\frac{x^2+4x}{(3x^3+2)^4}=\frac{u(x)}{v(x)}$ Use the quotient rule: $\frac{dy}{dx}=\frac{v(t).u'(t)-u(t).v'(t)}{[v(t)]^2}$ $\frac{dy}{dx}=\frac{(3x^3+2)^4(2x+4)-(x^2+4x)4(3x^3+2)^3.9x^2}{[(3x^3+2)^4]^2}$ $=\frac{(3x^3+2)^3[(3x^3+2)(2x+4)-36x^2(x^2+4x)]}{[(3x^3+2)^4]^2}$ $=\frac{(3x^3+2)^3[6x^4+4x+12x^3+8-36x^4-144x^3]}{(3x^3+2)^8}$ $=\frac{-30x^4+4x-132x^3+8}{(3x^3+2)^5}$