Question

attempt 1: 10 attempts remaining. suppose that $f(x) = \frac{9}{\ln(x^3 + 3)}$. find $f(2)$. $f(2) = $ submit answer

Explanation:

Step 1: Recall the quotient rule and derivative of ln

The function \( f(x)=\frac{9}{\ln(x^3 + 3)} \) can be written as \( f(x)=9(\ln(x^3 + 3))^{-1} \). We use the chain rule. The derivative of \( u^{-1} \) with respect to \( x \) (where \( u = \ln(x^3+3) \)) is \( -u^{-2}\cdot u' \). First, find the derivative of \( \ln(x^3 + 3) \). The derivative of \( \ln(v) \) with respect to \( x \) (where \( v=x^3 + 3 \)) is \( \frac{v'}{v} \), and \( v' = 3x^2 \). So the derivative of \( \ln(x^3 + 3) \) is \( \frac{3x^2}{x^3 + 3} \).

Step 2: Apply the chain rule to \( f(x) \)

Using the chain rule, \( f'(x)=9\times(-1)\times(\ln(x^3 + 3))^{-2}\times\frac{3x^2}{x^3 + 3} \)
Simplify it to \( f'(x)=-\frac{27x^2}{(\ln(x^3 + 3))^2(x^3 + 3)} \)

Step 3: Substitute \( x = 2 \) into \( f'(x) \)

First, calculate \( x^3+3 \) when \( x = 2 \): \( 2^3+3=8 + 3=11 \)
Then, calculate \( \ln(11) \), and \( (\ln(11))^2 \)
Also, \( x^2=4 \)
Substitute these values into \( f'(x) \):
\( f'(2)=-\frac{27\times4}{(\ln(11))^2\times11}=-\frac{108}{11(\ln(11))^2} \)
We can also calculate the numerical value. \( \ln(11)\approx2.3979 \), \( (\ln(11))^2\approx5.750 \)
\( 11\times5.750\approx63.25 \)
\( \frac{108}{63.25}\approx1.707 \)
So \( f'(2)\approx - 1.707 \) (or keep it in exact form \( -\frac{108}{11(\ln 11)^2} \))

Answer:

\( -\frac{108}{11(\ln 11)^2} \) (or approximately \( -1.71 \))