Help me find derivative of 4(x^(2/3) - 1) / 3x^(1/3)

[hank]

I need the derivative of 4(x^(2/3) - 1) / 3x^(1/3).

The answer I get is:

(4x^(-2/3))/3 + (4x^(-4/3)/3

But I don't think this is correct.

Thanks!

 

[galactus]

Hey Hankster.

Since you know how to differentiate, I will skip the formalities.

Its

\(\displaystyle \L\\\frac{4(x^{\frac{2}{3}}+1)}{9x^{\frac{4}{3}}}\)

 

[hank]

Hmm...Not sure if I put the problem up there right.

The closest I get to your answer is:

(x^(-2/3) -4) / 9x^(2/3) using the quotient rule.

Here's my steps:

d/dx[ (4(x^(2/3) - 1) / (3x^(1/3))]
= [(3x^(1/3) * (8/3)x^(-1/3) - 4(x^(2/3) - 1) * x^(-2/3)] / (3x^(1/3))^2
= (8 - 4 - x^(-2/3)) / 9x^(2/3)
= (x^(-2/3) -4) / 9x^(2/3)

 

[hank]

Got it...took me a couple hours, but finally got it.

 

[galactus]

\(\displaystyle \L\\\frac{(3x^{\frac{1}{3}})(\frac{8}{3x^{\frac{1}{3}}})-(4x^{\frac{2}{3}}-4)(x^{\frac{-2}{3}})}{9x^{\frac{2}{3}}}\)

=\(\displaystyle \L\\\frac{8-(4x^{\frac{2}{3}}-4)x^{\frac{-2}{3}}}{9x^{\frac{2}{3}}}\)

=\(\displaystyle \L\\\frac{4(x^{\frac{2}{3}}+1)}{9x^{\frac{4}{3}}}\)