How do you find the derivative of #(x^(2/3 ))(8-x)#?

Question

Charles Anctil · Accepted Answer

#( d (8x^(2/3) - x^(-1/3)))/(dx) = 1/(3root(3)(x)) *(16+1/x)# #(x^(2/3))(8-x)# #color(white)("XXXX")##=8x^(2/3) - x^(-1/3)# #( d (8x^(2/3) - x^(-1/3)))/(dx) = color(red)((d (8x^(2/3)))/(dx)) - color(blue)((d (x^(-1/3)))/(dx))# #color(white)("XXXX")##=color(red)( 2/3*8x^-(1/3)) - color(blue)((-1/3)x^(-4/3))# #color(white)("XXXX")##= 16/(3root(3)x) + 1/(3(root(3)(x))^4)# #color(white)("XXXX")##= 16/(3root(3)x) + 1/(3(x*root(3)(x)))# #color(white)("XXXX")##= 1/(3root(3)(x)) *(16+1/x)#

Scarlett Allison · Answer

undefined To find the derivative of $ (x^\frac{2}{3})(8-x) $, apply the product rule, which states that the derivative of the product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.

So, differentiate $ x^\frac{2}{3} $ with respect to $ x $ to get $ \frac{2}{3}x^{-\frac{1}{3}} $, and differentiate $ (8-x) $ with respect to $ x $ to get $ -1 $. Then apply the product rule:

$$ \frac{d}{dx} \left( x^\frac{2}{3} \cdot (8-x) \right) = \frac{2}{3}x^{-\frac{1}{3}} \cdot (8-x) + x^\frac{2}{3} \cdot (-1) $$

Simplify the expression to get the derivative:

$$ \frac{d}{dx} \left( x^\frac{2}{3} \cdot (8-x) \right) = \frac{16}{3}x^{-\frac{1}{3}} - x^\frac{2}{3} $$