Not understanding this derivative

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Pikachu!

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It's for a global max/min problem, and I understand those concepts, but I need the derivative for this in order to get the critical number(s) to solve the whole problem.


f(x)=x(x-1)1/3 I know the answer is (4x-3)/3(x-1)2/3
My first thought was to simply distribute the outside x to make it (x^2-x)1/3 but that led me to an incorrect answer of 2x-1/3(x^2-x)2/3

I honestly just have no idea how to arrive at that answer, but I've checked multiple sources and I know it is the right one, because I worked out the resulting critical number into my global min/max problem, and it came out to the answer given in the book.

I feel like my error is somewhere in my algebra more than anything, but I really just have no idea and could really use some help. This class is killing me.

Thank you so much.
 
Distribute...really? That should work, but you must think it through a little better.

\(\displaystyle x(x-1)^{1/3} = (x^{3}(x-1))^{1/3} = (x^{4}-x^{3})^{1/3}\)

Give that a try.
 
tkhunny said:
Distribute...really? That should work, but you must think it through a little better.

\(\displaystyle x(x-1)^{1/3} = (x^{3}(x-1))^{1/3} = (x^{4}-x^{3})^{1/3}\)

Give that a try.
Click to expand...

Er, how?

x(x-1)1/3 wouldn't turn into (x^2-x)1/3?
 
Pikachu! said:
Er, how?

x(x-1)1/3 wouldn't turn into (x^2-x)1/3?
Click to expand...
For goodness sake, you can't just take a a variable unchanged into an expression raised to a power. Try an example:

\(\displaystyle 9 * \left(9 - 1\right)^{(1/3)} = 9\sqrt[3]{8} = 9 * 2 = 18 \ne\)

\(\displaystyle \left\{9(9 - 1)\right\}^{(1/3)} =\sqrt[3]{9 * 8} = 2\sqrt[3]{9} \approx 4.16.\)

\(\displaystyle x = \left(x^3\right)^{(1/3)}.\)
 
Pikachu! said:
This class is killing me.

Click to expand...


You are in a calculus class, and you do not understand the Order of Operations?

Have you been out of school for a long time?

I think the reason why you feel that your class is killing you is because you're in the wrong class!
 
Hi Pikachu

The reason that mmm4444bot suggested that you think about the Order of Operations is because you interpreted the problem as ...

f(x)=(x(x-1))1/3

If the problem had been written that way -- you would be correct.
However the problem was written as ...

f(x)=x(x-1)1/3

It is a common mistake, even for a Calculus student. What is not common for a Calculus student is to not recognize this kind of mistake when it is pointed out. You know your situation. We could be wrong, and certainly do not intend to insult you.

If you think this is a weakness that you agree is keeping you from enjoying Calculus, you may want to consider taking a course or two of Algebra over again.

If there is anyway that we can help you be successful, be sure to post on FMH.
Look forward to hearing from you :)
 
Pikachu! said:
f(x)=x(x-1)1/3 I know the answer is (4x-3)/3(x-1)2/3

Thank you so much.
Click to expand...

No, that is not the answer.


What you typed is equivalent to:


\(\displaystyle \bigg(\dfrac{4x - 3}{3}\bigg)(x - 1)^{2/3}, \ \ due \ \ to \ \ the \ \ Order \ \ of \ \ Operations.\)


And you left out the "f'(x) = " portion.



Place some type of matching pair of grouping symbols in the appropriate part of the denominator:


f'(x) = (4x - 3)/(3(x - 1)2/3) or


f'(x) = (4x - 3)/[3(x - 1)2/3] or


f'(x) = (4x - 3)/{3(x - 1)2/3}

 
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