What is the derivative of #(x^(2/3 ))(8-x)#?

1 Answer
Johnson Z.
Feb 28, 2016

#[(x^(2/3))(8-x)]'=(16-5x)/(3root(3)(x))#

Explanation:

Recall the following:

#1#. Product Rule: #[f(x)g(x)]'=f(x)g'(x)+f'(x)g(x)#
#2#. Difference Rule: #[f(x)-g(x)]'=f'(x)-g'(x)#
#3#. Constant Rule: #c'=0#
#4#. Power Rule: #(x^n)'=nx^(n-1)#

Finding the Derivative
#1#. Start by rewriting the equation using the product rule.

#[(x^(2/3))(8-x)]'#

#=(x^(2/3))(color(red)(8-x))'+(color(purple)(x^(2/3)))'(8-x)#

#2#. Use the difference rule and constant rule for #(color(red)(8-x))#. Recall that the derivative of a constant is always #color(orange)0# and the derivative of #x# is always #color(blue)1#.

#=(x^(2/3))(color(red)(8'-x'))+(color(purple)(x^(2/3)))'(8-x)#

#=(x^(2/3))(color(orange)0-color(blue)1)+(color(purple)(x^(2/3)))'(8-x)#

#3#. Use the power rule for #color(purple)(x^(2/3))#.

#=(x^(2/3))(0-1)+(2/3x^(-1/3))(8-x)#

#4#. Simplify.

#=-x^(2/3)+(2/(3x^(1/3)))(8-x)#

#=(2/(3root(3)(x)))(8-x)-x^(2/3)#

#=(16-2x)/(3root(3)(x))-x^(2/3)#

#=(16-2x)/(3root(3)(x))-(x^(2/3)(3root(3)(x)))/(3root(3)(x))#

#=(16-2x-x^(2/3)(3root(3)(x)))/(3root(3)(x))#

#=(16-2x-x^(2/3)(3x^(1/3)))/(3root(3)(x))#

#=(16-2x-3x)/(3root(3)(x))#

#=color(green)((16-5x)/(3root(3)(x)))#

#:.#, the derivative is #(16-5x)/(3root(3)(x))#.

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