3.3c differentiation ru\n(6 points)\nlet $f(x)=\\frac{9x^{2}}{6 - 5x}$. then\n$f(x)=\\square$\nand\n$f(x)=\\s…

3.3c differentiation ru\n(6 points)\nlet $f(x)=\\frac{9x^{2}}{6 - 5x}$. then\n$f(x)=\\square$\nand\n$f(x)=\\square$\nnote: you can earn partial credit on this problem.\nnote: you are in the reduced scoring period. all work co\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have 5 attempts remaining.

3.3c differentiation ru\n(6 points)\nlet $f(x)=\\frac{9x^{2}}{6 - 5x}$. then\n$f(x)=\\square$\nand\n$f(x)=\\square$\nnote: you can earn partial credit on this problem.\nnote: you are in the reduced scoring period. all work co\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have 5 attempts remaining.

Answer

Explanation:

Step1: Apply the quotient rule for (f'(x))

The quotient rule is (\left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}}). Here (u = 9x^{2}), (u'=18x), (v = 6 - 5x), (v'=- 5). [ \begin{align*} f'(x)&=\frac{(18x)(6 - 5x)-9x^{2}(-5)}{(6 - 5x)^{2}}\ &=\frac{108x-90x^{2}+45x^{2}}{(6 - 5x)^{2}}\ &=\frac{108x - 45x^{2}}{(6 - 5x)^{2}} \end{align*} ]

Step2: Apply the quotient rule for (f''(x))

Now, for (y = f'(x)=\frac{u}{v}) where (u = 108x - 45x^{2}), (u'=108 - 90x), (v=(6 - 5x)^{2}), and using the chain - rule (v' = 2(6 - 5x)(-5)=-10(6 - 5x)) [ \begin{align*} f''(x)&=\frac{(108 - 90x)(6 - 5x)^{2}-(108x - 45x^{2})(-10)(6 - 5x)}{(6 - 5x)^{4}}\ &=\frac{(6 - 5x)[(108 - 90x)(6 - 5x)+10(108x - 45x^{2})]}{(6 - 5x)^{4}}\ &=\frac{108\times6-108\times5x-90x\times6 + 450x^{2}+1080x-450x^{2}}{(6 - 5x)^{3}}\ &=\frac{648-540x-540x + 450x^{2}+1080x-450x^{2}}{(6 - 5x)^{3}}\ &=\frac{648}{(6 - 5x)^{3}} \end{align*} ]

Answer:

(f'(x)=\frac{108x - 45x^{2}}{(6 - 5x)^{2}}), (f''(x)=\frac{648}{(6 - 5x)^{3}})