Question

suppose that $f(x)=\frac{9}{ln(x^{3}+2)}$. find $f(3).$ $f(3)=$

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = 9$, so $u'=0$, and $v=\ln(x^{3}+2)$, so $v'=\frac{3x^{2}}{x^{3}+2}$ by the chain - rule. Then $f'(x)=\frac{0\times\ln(x^{3}+2)-9\times\frac{3x^{2}}{x^{3}+2}}{(\ln(x^{3}+2))^{2}}=-\frac{27x^{2}}{(x^{3}+2)(\ln(x^{3}+2))^{2}}$.

Step2: Substitute $x = 3$

Substitute $x = 3$ into $f'(x)$. First, calculate $x^{3}+2$ when $x = 3$: $3^{3}+2=27 + 2=29$. Then calculate $\ln(x^{3}+2)=\ln(29)$. Now, $f'(3)=-\frac{27\times3^{2}}{29\times(\ln(29))^{2}}=-\frac{27\times9}{29\times(\ln(29))^{2}}=-\frac{243}{29(\ln(29))^{2}}$.

Answer:

$-\frac{243}{29(\ln(29))^{2}}$