To find the equation of the tangent line to the curve $y = x^{1/3} - \frac{16}{x}$ at the point $(8, 0)$, follow these steps:

Step 1: Find the derivative of $y$ with respect to $x$

The equation of the tangent line requires the slope, which is found by differentiating $y$ with respect to $x$.

$y = x^{1/3} - \frac{16}{x}$

Apply the power rule and quotient rule to differentiate:

1. The derivative of $x^{1/3}$: $\frac{d}{dx} x^{1/3} = \frac{1}{3}x^{-2/3}$

2. The derivative of $-\frac{16}{x}$ (which is $-16x^{-1}$): $\frac{d}{dx} \left( -16x^{-1} \right) = 16x^{-2}$

Thus, the derivative of $y$ is:

$\frac{dy}{dx} = \frac{1}{3}x^{-2/3} + 16x^{-2}$

Step 2: Evaluate the derivative at $x = 8$

Now, substitute $x = 8$ into the derivative to find the slope of the tangent line at the point $(8, 0)$.

$\frac{dy}{dx} \Bigg|_{x=8} = \frac{1}{3}(8)^{-2/3} + 16(8)^{-2}$

First, simplify each term:

1. $8^{-2/3} = \frac{1}{4}$ (since $8^{1/3} = 2$ and $2^2 = 4$),
2. $8^{-2} = \frac{1}{64}$.

So, the slope is:

$\frac{dy}{dx} \Bigg|_{x=8} = \frac{1}{3} \times \frac{1}{4} + 16 \times \frac{1}{64} = \frac{1}{12} + \frac{1}{4} = \frac{1}{12} + \frac{3}{12} = \frac{4}{12} = \frac{1}{3}$

Step 3: Find the equation of the tangent line

The slope of the tangent line at $(8, 0)$ is $\frac{1}{3}$. Using the point-slope form of the equation of a line:

$y - y_1 = m(x - x_1)$

where $m = \frac{1}{3}$, $x_1 = 8$, and $y_1 = 0$, we get:

$y - 0 = \frac{1}{3}(x - 8)$

Simplify:

$y = \frac{1}{3}(x - 8)$

Thus, the equation of the tangent line is:

$y = \frac{1}{3}x - \frac{8}{3}$

Final Answer:

The equation of the tangent line to the curve $y = x^{1/3} - \frac{16}{x}$ at the point $(8, 0)$ is:

$y = \frac{1}{3}x - \frac{8}{3}$

Would you like more details on any of the steps, or do you have any questions?

Here are 5 related questions:

1. How do you find the tangent line to a curve at a general point?
2. What is the power rule of differentiation, and how does it apply here?
3. How would the process change if the curve had different exponents?
4. Can you find the tangent line if the point lies off the curve?
5. How does the quotient rule apply to differentiating functions like $-\frac{16}{x}$?

Tip: When finding derivatives of fractional powers, rewriting terms with negative exponents often simplifies the differentiation process.