# What is the derivative of (6)/(x^3sqrtx)?

The answer is: $y ' = - \frac{21}{{x}^{4} \sqrt{x}}$.
$y = 6 \cdot {x}^{- 3 - \frac{1}{2}} \Rightarrow y = 6 {x}^{- \frac{7}{2}}$
$y ' = 6 \cdot \left(- \frac{7}{2}\right) \cdot {x}^{- \frac{7}{2} - 1} = - 21 {x}^{- \frac{9}{2}} = - \frac{21}{\sqrt{{x}^{9}}} =$
$= - \frac{21}{{x}^{4} \sqrt{x}}$.