# Math Help - Exponent problem in quotent rule

1. ## Exponent problem in quotent rule

I'm trying to solve a quotient problem but apparently my radical exponent algebra is rusty. Can someone tell me if this is right?

use quotent rule to differentiate the function:
(x^(1/3))/(x^3+1)

I've used the rule to get:
(x^3+1) 1/3 x^(-2/3) - x^(1/3) (3x^2) / (x^3+1)^2

My problem is the
1/3 x^(-2/3) - x^(1/3) (3x^2) (I'll plug this back in)

I transform into 1/
3x^(2/3) - (x^(1/3) (3x^2) (3x^(2/3))) / 3x^(2/3)

then
1- (x^(1/3) (3x^(2/1)) (3x^(2/3))) / 3x^(2/3)

then 1 - (x^(1/3)9x^(8/3)) / 3x^(2/3)

then 1 - 9x^3 / 3x^(2/3)

plug back in to get: x^3 +1 - 9x^3 / 3x^(2/3) (x^3+1)^2

then 1-8
x^3 / 3x^(2/3) (x^3+1)^2

That is the correct answer but i just want to make sure i am using the radical multiplication rule correctly.
Thanks.

2. Originally Posted by ixo
I'm trying to solve a quotient problem but apparently my radical exponent algebra is rusty. Can someone tell me if this is right?

use quotent rule to differentiate the function:
(x^(1/3))/(x^3+1)

I've used the rule to get:
(x^3+1) 1/3 x^(-2/3) - x^(1/3) (3x^2) / (x^3+1)^2

My problem is the
1/3 x^(-2/3) - x^(1/3) (3x^2) (I'll plug this back in)

I transform into 1/
3x^(-2/3) - (x^(1/3) (3x^2) (3x^(2/3))) / 3x^(2/3)

then
1- (x^(1/3) (3x^(2/1)) (3x^(2/3))) / 3x^(2/3)

then 1 - (x^(1/3)9x^(8/3)) / 3x^(2/3)

then 1 - 9x^3 / 3x^(2/3)

plug back in to get: x^3 +1 - 9x^3 / 3x^(2/3) (x^3+1)^2

then( 1-8
x^3 )/( 3x^(2/3) (x^3+1)^2)

That is the correct answer but i just want to make sure i am using the radical multiplication rule correctly.
Thanks.
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1. Your calculations are correct if I add some brackets to separate numerator and denominator of the quotient.

2. You can abbreviate your calculations here
(x^(1/3) (3x^2) (3x^(2/3))) / 3x^(2/3)
a little bit if you change the order of the factors:

$\dfrac{x^{\frac13} \cdot 3x^2 \cdot 3x^{\frac23}}{3x^{\frac23}} = \dfrac{ 3x^2 \cdot 3x^{\frac23} \cdot x^{\frac13} }{3x^{\frac23}} = \dfrac{ 3x^2 \cdot 3x^{\frac23+\frac13} }{3x^{\frac23}} = \dfrac{3x^2 \cdot 3x}{3x^{\frac23}}$