derivative equation

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k I'm having trouble with this question, could someone just show me all the steps to this as it's a tricky one.

y=[(x+2)/ (x^1/3)]^3 What is the derivative to this?

thanks for the help in advance
 
bittersweet said:
I'm having trouble with this question....
Click to expand...
Please reply showing your steps and stating where you're having the trouble, and the tutors will be glad to check over your work.

Thank you.

Eliz.
 
Sometimes a little Algebra is useful.

\(\displaystyle \L\begin{array}{l}
y = \left( {\frac{{x + 2}}{{\sqrt[3]{x}}}} \right)^3 \\
y = \frac{{(x + 2)^3 }}{x} \\
\end{array}\)

I think that is fairly easy to differentiate using quotient rule.
If you prefer, you could expand the numerator, divide every term by x, and then it is a simple power rule problem.

Please post your solution to the problem.
 
sorry where did the x^1/3 on the denominator go? I'm having trouble seeing where that went..
 
What does \(\displaystyle (x^{\frac{1}{3}})^{3}\) equal?.
 
ok I did something wrong, y=(x+2)^3/x
y'(x)= 3(x+2)^2(1)(x)-[(x+2)^3(1)]/x^2
y'(x)= 3x(x+2)^2-(x+2)^3/x^2
y'(x)= (x+2)^2(3x)-(x+2)/x^2

what part did I mess up on?
 
bittersweet said:
ok I did something wrong, y=(x+2)^3/x
y'(x)= 3(x+2)^2(1)(x)-[(x+2)^3(1)]/x^2
y'(x)= 3x(x+2)^2-(x+2)^3/x^2
y'(x) = [(x+2)^2][3x - (x+2)]/x^2
y'(x) = [(x+2)^2](2x - 2)/x^2
y'(x) = [2(x - 1)(x + 2)^2]/x^2

what part did I mess up on? no place, just finish up
Click to expand...

the algebra is killin' you, isn't it?
 
I guess so,do you have a site for algerbra review :p cuz our teacher says the algerbra usually kills all since it's stuff form like back gr.9 and she told us to review our gr.9 notes..cept that's gr 9!! =p I don't have old notes ^_^ and Thank you
 
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