Get the answers you need at Westonci.ca, where our expert community is always ready to help with accurate information. Get the answers you need quickly and accurately from a dedicated community of experts on our Q&A platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
[tex]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)[/tex]
[tex]\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}[/tex]
now, we could expand the polynomials, but there isn't much simplification, so no much point doing so.
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.
