Differentiate (8 + x) / (2 + x^(1/3))^2

[hank]

Ok, here we go...

f'(x) = [(2 + x^(1/3)) - (8 + x)(1/3)(x^(-2/3)) ] / (x + x^(1/3))^2

Not sure where to go from here.

I can distribute the (1/3)(x^(-2/3)), but I'm not sure what that buys me, if anything.

Help?

 

[galactus]

Try chnaging it to the product rule. May be easier. I always do that if I can.

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

 

[hank]

Ok, I gave that a try and here's what I got...

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

//Product rule
= [(8 + x)(-2)(2 + x^(1/3)^-3((1/3)x^(-2/3)] + (2 + x^(1/3))^-2

//Multiplying out and combining
=[(-x - 16)/(3x^(2/3)(2 + x^(1/3))^3)] + [1/(2 + x^(1/3))^2]

//Find common denominator and add
((6x^(2/3) - 16) / (3x^(2/3)(2+x^(1/3))^3)

\(\displaystyle \frac{{{\rm 6x}^{\frac{{\rm 2}}{{\rm 3}}} {\rm - 16}}}{{{\rm 3x}^{\frac{{\rm 2}}{{\rm 3}}} {\rm (2 + x}^{\frac{{\rm 1}}{{\rm 3}}} {\rm )}^{\rm 3} }}
\\)

Does that look right, or can I do more?

 

[galactus]

Ok, I gave that a try and here's what I got...

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

//Product rule
= [(8 + x)(-2)(2 + x^(1/3)^-3((1/3)x^(-2/3)] + (2 + x^(1/3))^-2

//Multiplying out and combining
=[(-x - 16)/(3x^(2/3)(2 + x^(1/3))^3)] + [1/(2 + x^(1/3))^2]

//Find common denominator and add
((6x^(2/3) - 16) / (3x^(2/3)(2+x^(1/3))^3)

Does that look right, or can I do more?

You forgot an x in the numerator. That's all.

\(\displaystyle \L\\\frac{x+6x^{\frac{2}{3}}-16}{3x^{\frac{2}{3}}(x^{\frac{1}{3}}+2)^{3}}\)

 

[hank]

Boy, I feel silly...

The problem was actually:
\(\displaystyle \frac{{8 + x}}{{2 + \sqrt[3]{x}}}
\\)

Here's the answer I got:

\(\displaystyle \frac{{{\rm - x}^{\frac{{\rm 4}}{{\rm 3}}} - 8x^{\frac{1}{3}} - 18}}{{3x^{\frac{2}{3}} \left( {2 + x^{\frac{1}{3}} } \right)}}\)

Does that look right?
Also, can I factor the top somehow?