# How do you simplify (x^(-1/2)y^2)^(-5/4)/(x^2y^(1/2)?

May 3, 2017

$f \left(x , y\right) = \frac{1}{{x}^{\frac{11}{8}} {y}^{3}}$

#### Explanation:

Given:
$f \left(x , y\right) = \frac{{\left({x}^{- \frac{1}{2}} {y}^{2}\right)}^{- \frac{5}{4}}}{{x}^{2} {y}^{\frac{1}{2}}}$

We first distribute the power in the numerator using the Power Rule for exponents, since the base terms $x$ and $y$ are being multiplied together and not added.

$f \left(x , y\right) = \frac{{x}^{\left(- \frac{1}{2}\right) \left(- \frac{5}{4}\right)} {y}^{\left(2\right) \left(- \frac{5}{4}\right)}}{{x}^{2} {y}^{\frac{1}{2}}} = \frac{{x}^{\frac{5}{8}} {y}^{- \frac{5}{2}}}{{x}^{2} {y}^{\frac{1}{2}}}$

Now we apply the Quotient Rule for Exponents on both $x$ and $y$, following:

${x}^{a} / {x}^{b} = {x}^{a - b}$

$f \left(x , y\right) = {x}^{\frac{5}{8} - 2} {y}^{- \frac{5}{2} - \frac{1}{2}} = {x}^{- \frac{11}{8}} {y}^{- 3}$

Now, we apply the negative exponent rule,

${x}^{- a} = \frac{1}{x} ^ \left(a\right)$

$f \left(x , y\right) = \frac{1}{{x}^{\frac{11}{8}} {y}^{3}}$