200k views
3 votes
Find the derivative f(x)=[(3x^2-2)/(2x+3)]^3

User JoniJnm
by
8.5k points

1 Answer

4 votes


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.

User XavierB
by
7.6k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories