How do you use the quotient rule to find the derivative of #y=(x-sqrt(x))/(x^(1/3))# ?

Question

Ezekiel Alexander · Accepted Answer

#y'=2/3*1/x^(1/3)-1/6*1/x^(5/6)# Solution: #y=(x−sqrtx)/x^(1/3)# #y=x/x^(1/3)-x^(1/2)/x^(1/3)# #y=x^(2/3)-x^(1/6)# differentiating with respect to #x#, #y'=2/3x^(2/3-1)-1/6x^(1/6-1)# #y'=2/3x^(-1/3)-1/6x^(-5/6)# #y'=2/3*1/x^(1/3)-1/6*1/x^(5/6)#

Aurora Alvarado · Answer

undefined To find the derivative of $ y = \frac{x - \sqrt{x}}{x^{1/3}} $ using the quotient rule:

1. Identify $ u $ and $ v $ as the numerator and denominator functions respectively.
   $ u = x - \sqrt{x} $
   $ v = x^{1/3} $

2. Compute $ u' $ and $ v' $ (the derivatives of $ u $ and $ v $ respectively).
   $ u' = 1 - \frac{1}{2\sqrt{x}} $
   $ v' = \frac{1}{3}x^{-2/3} $

3. Apply the quotient rule: 
   $ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} $

4. Substitute $ u' $, $ v' $, $ u $, and $ v $ into the formula and simplify.

$$ \frac{d}{dx}\left(\frac{x - \sqrt{x}}{x^{1/3}}\right) = \frac{(1 - \frac{1}{2\sqrt{x}})x^{1/3} - (x - \sqrt{x})(\frac{1}{3}x^{-2/3})}{(x^{1/3})^2} $$

$$ = \frac{(x^{1/3} - \frac{1}{2}x^{-1/6}) - (x^{1/3} - x^{1/6})}{x^{2/3}} $$

$$ = \frac{x^{1/3} - \frac{1}{2}x^{-1/6} - x^{1/3} + x^{1/6}}{x^{2/3}} $$

$$ = \frac{- \frac{1}{2}x^{-1/6} + x^{1/6}}{x^{2/3}} $$

$$ = \frac{-\frac{1}{2}x^{-1/6} + x^{1/6}}{x^{2/3}} $$