We are given the following conditions:

• $f(8) = 2$
• $f'(8) = 5$
• The function $g(x)$ is defined as $g(x) = \frac{f(x)}{\sqrt[3]{x}}$

We are asked to find $g'(8)$.

Step 1: Use the Quotient Rule for Derivatives

To differentiate $g(x) = \frac{f(x)}{\sqrt[3]{x}}$, we will use the quotient rule, which states:

$\left( \frac{u(x)}{v(x)} \right)' = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$

Here, we identify $u(x) = f(x)$ and $v(x) = \sqrt[3]{x} = x^{1/3}$.

Now, we need the derivatives of both $u(x)$ and $v(x)$.

Step 2: Find the Derivatives

• The derivative of $u(x) = f(x)$ is $u'(x) = f'(x)$.
• The derivative of $v(x) = x^{1/3}$ is:

$v'(x) = \frac{1}{3} x^{-2/3}$

Step 3: Apply the Quotient Rule

Now, applying the quotient rule:

$g'(x) = \frac{f'(x) \cdot x^{1/3} - f(x) \cdot \frac{1}{3} x^{-2/3}}{(x^{1/3})^2}$

Simplifying the denominator:

$(x^{1/3})^2 = x^{2/3}$

Thus:

$g'(x) = \frac{f'(x) \cdot x^{1/3} - \frac{f(x)}{3} \cdot x^{-2/3}}{x^{2/3}}$

Step 4: Evaluate at $x = 8$

Now, substitute $x = 8$, along with the given values $f(8) = 2$ and $f'(8) = 5$.

First, calculate the powers of 8:

$8^{1/3} = 2 \quad \text{and} \quad 8^{-2/3} = \frac{1}{4}$

Now, substitute into the expression for $g'(x)$:

$g'(8) = \frac{5 \cdot 2 - \frac{2}{3} \cdot \frac{1}{4}}{8^{2/3}} = \frac{10 - \frac{2}{12}}{4} = \frac{10 - \frac{1}{6}}{4}$

Simplifying:

$g'(8) = \frac{\frac{60}{6} - \frac{1}{6}}{4} = \frac{\frac{59}{6}}{4} = \frac{59}{24}$

Thus, the value of $g'(8)$ is:

$g'(8) = \frac{59}{24}$

Would you like more details or further explanation on any part of this solution?

Related Questions:

1. How does the chain rule apply when differentiating a composition of functions?
2. Can the quotient rule be simplified for specific types of functions?
3. What are other common applications of the cube root in calculus problems?
4. How does one compute limits of functions involving cube roots?
5. How would the result differ if $v(x)$ had a higher or lower exponent?

Tip:

When applying the quotient rule, always simplify the result by factoring out common terms if possible to make the expression easier to evaluate.