Question

Find the derivative f(x)=[(3x^2-2)/(2x+3)]^3

Answer 1

`f(x)=\left( \cfrac{3x^2-2}{2x+3} \right)^3\implies \cfrac{df}{dx}=\stackrel{\textit{\LARGE chain~ ~~ rule}}{3\left( \cfrac{3x^2-2}{2x+3} \right)^2\underset{quotient~rule}{\left( \cfrac{6x(2x+3)~~ - ~~(3x^2-2)2}{(2x+3)^2} \right)}} \\\\\\ \cfrac{df}{dx}=3\left( \cfrac{3x^2-2}{2x+3} \right)^2\left( \cfrac{12x^2+18x-6x^2+4}{(2x+3)^2} \right) \\\\\\ \cfrac{df}{dx}=3\left( \cfrac{3x^2-2}{2x+3} \right)^2\left( \cfrac{6x^2+18x+4}{(2x+3)^2} \right)`

`\cfrac{df}{dx}=3 \cfrac{(3x^2-2)^2}{(2x+3)^2} \left( \cfrac{2(3x^2+9x+2)}{(2x+3)^2} \right)\implies \cfrac{df}{dx}=\cfrac{6(3x^2-2)^2(3x^2+9x+2)}{(2x+3)^4}`

now, we could expand the polynomials, but there isn't much simplification, so no much point doing so.