The problem is to simplify the expression: $(\frac{-2a^2b^{\frac{3}{4}}}{12a^3b^{\frac{1}{2}}})^2$

1. Problem Description

The problem is to simplify the expression:

$(\frac{-2a^2b^{\frac{3}{4}}}{12a^3b^{\frac{1}{2}}})^2$

2. Solution Steps

First, simplify the expression inside the parenthesis:

$\frac{-2a^2b^{\frac{3}{4}}}{12a^3b^{\frac{1}{2}}} = \frac{-1}{6} \cdot \frac{a^2}{a^3} \cdot \frac{b^{\frac{3}{4}}}{b^{\frac{1}{2}}}$

Using the quotient rule for exponents, $\frac{x^m}{x^n} = x^{m-n}$:

$\frac{a^2}{a^3} = a^{2-3} = a^{-1}$

$\frac{b^{\frac{3}{4}}}{b^{\frac{1}{2}}} = b^{\frac{3}{4}-\frac{1}{2}} = b^{\frac{3}{4}-\frac{2}{4}} = b^{\frac{1}{4}}$

So, the expression inside the parenthesis becomes:

$\frac{-1}{6}a^{-1}b^{\frac{1}{4}}$

Now, we raise the entire expression to the power of 2:

$(\frac{-1}{6}a^{-1}b^{\frac{1}{4}})^2 = (\frac{-1}{6})^2 (a^{-1})^2 (b^{\frac{1}{4}})^2$

Using the power of a product rule, $(xy)^n = x^n y^n$ and the power of a power rule, $(x^m)^n = x^{mn}$:

$(\frac{-1}{6})^2 = \frac{1}{36}$

$(a^{-1})^2 = a^{-2}$

$(b^{\frac{1}{4}})^2 = b^{\frac{2}{4}} = b^{\frac{1}{2}}$

Therefore, the expression simplifies to:

$\frac{1}{36}a^{-2}b^{\frac{1}{2}}$

We can rewrite $a^{-2}$ as $\frac{1}{a^2}$, so the final simplified expression is:

$\frac{1}{36} \cdot \frac{1}{a^2} \cdot b^{\frac{1}{2}} = \frac{b^{\frac{1}{2}}}{36a^2}$

3. Final Answer

$\frac{b^{\frac{1}{2}}}{36a^2}$