Question

hw_3.1_the chain rule
turn in assignment
due sep 21, 2025 11:59 pm.

1. submit answer practice similar

attempt 6: 5 attempts remaining.
suppose that $f(x)=\frac{8}{ln(x^{3}+3)}$. find $f(3)$.
$f(3)=$

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = 8$ and $v=\ln(x^{3}+3)$. So, $u' = 0$ and we need to find $v'$ using the chain - rule.

Step2: Find $v'$ using chain - rule

Let $t=x^{3}+3$. Then $v = \ln(t)$. By the chain - rule, $\frac{dv}{dx}=\frac{dv}{dt}\cdot\frac{dt}{dx}$. Since $\frac{d}{dt}(\ln(t))=\frac{1}{t}$ and $\frac{d}{dx}(x^{3}+3)=3x^{2}$, we have $v'=\frac{3x^{2}}{x^{3}+3}$.

Step3: Apply quotient - rule to find $f'(x)$

$f'(x)=\frac{0\times\ln(x^{3}+3)-8\times\frac{3x^{2}}{x^{3}+3}}{(\ln(x^{3}+3))^{2}}=-\frac{24x^{2}}{(x^{3}+3)(\ln(x^{3}+3))^{2}}$.

Step4: Evaluate $f'(3)$

Substitute $x = 3$ into $f'(x)$. When $x = 3$, $x^{3}+3=3^{3}+3=27 + 3=30$ and $\ln(x^{3}+3)=\ln(30)$. Then $f'(3)=-\frac{24\times3^{2}}{30\times(\ln(30))^{2}}=-\frac{24\times9}{30\times(\ln(30))^{2}}=-\frac{72}{10\times(\ln(30))^{2}}=-\frac{36}{5(\ln(30))^{2}}$.

Answer:

$-\frac{36}{5(\ln(30))^{2}}$