To solve the given limit problem:

$\lim_{{h \to 0}} \frac{{(16 + h)^{\frac{1}{4}} - 2}}{h}$

we can use the derivative definition. Let's denote $f(x) = x^{\frac{1}{4}}$. Then the derivative $f'(x)$ at $x = 16$ is given by:

$f'(16) = \lim_{{h \to 0}} \frac{{f(16 + h) - f(16)}}{h}$

We know that $f(16) = 2$, so we have:

$f'(16) = \lim_{{h \to 0}} \frac{{(16 + h)^{\frac{1}{4}} - 2}}{h}$

Now, we can find $f'(x)$ in general. Using the power rule:

$f'(x) = \frac{1}{4} x^{-\frac{3}{4}}$

At $x = 16$:

$f'(16) = \frac{1}{4} \times 16^{-\frac{3}{4}}$

Calculating $16^{-\frac{3}{4}}$:

$16 = 2^4 \quad \text{so} \quad 16^{\frac{1}{4}} = 2 \quad \text{and} \quad 16^{-\frac{3}{4}} = 2^{-3} = \frac{1}{8}$

Thus:

$f'(16) = \frac{1}{4} \times \frac{1}{8} = \frac{1}{32}$

The correct answer is:

Option B: $\frac{1}{32}$

Would you like more details on this solution or any further questions?

Here are some related questions:

1. How do we find the derivative of a function using the definition of limits?
2. Why does $(16 + h)^{\frac{1}{4}}$ approximate to 2 when $h$ is close to 0?
3. How do we compute the derivative of a function of the form $x^{\frac{1}{n}}$?
4. What is the geometric interpretation of a derivative at a point?
5. Can the limit definition of the derivative be used for all functions?

Tip: When working with limits involving roots, it can be helpful to rationalize the expression to simplify calculations.