Super Easy Derivative that has given me a headache

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Super Easy Derivative that has given me a headache!

Homework Statement



So I have to get the derivative for f(x) = x^3/1 *(x-2)
I have to find:
a) Critical Numbers
b) Intervals where it decreases/ increases
c) Local extremes
d) Concave intervals
Then the second derivative of that to get the concave intervals

The Attempt at a Solution



I tried doing the second derivative but I seem to get a wrong answer everytime.I get f(x) = 4x-2 / 3x^2/3 for the first derivative

But I get (12x^2/3) / x^4/3 + 4^x1/3

Can somebody help? I just need to know how to get the second derivative... And no I'm not asking for someone to do my Homework.

Edit: I will post my procedure

f(x)= x^1/3(x-2)
f'(x)= (1/3x^-2/3)(x-2) + (x^1/3)(1) Here I used the product rule (f*g)' = f'*g + f*g'
f'(x)= x-2/3x^2/3 + x^1/3
f'(x)= x-2/3x^2/3 + x^1/3*(3x^2/3)/(3x^2/3) I multiply by (3x^2/3)/(3x^2/3) to make it equal
f'(x)= x-2+3x/ 3x^2/3
f'(x)= 4x-2/3x^2/3

f'(x)=4x-2/3x^2/3
f''(x)= (4)(3x^2/3) - (2^-1/3)(4x-2) / (3x^2/3)^2 Here I use the quotient rule f'(x)= f'*g - g'*f / g^2
f''(x)= 12x^2/3 - (8x^-4/3 - 4x^-1/3) / 9x^4/3
f''(x)= 12x^2/3 / 9x^4/3 - 8x^4/3 + 4x^1/3
f''(x)= 12x^2/3 / x^4/3 + 4x^1/3
 
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Please use LaTeX (or at least parentheses) to correctly format your question. At the moment it is not clear what you mean.

Is x^3/1 *(x-2)

x^3(x-2)

or

\frac {x^3} {x-2}

or something else?
 


I would love to use LaTex but how can I use it to explain myself better?

Edit: I posted my entire procedure I hope it's clear. No it's none of those its the cubic root of x (not x^2). or (x^2/3)

Edit 2: I don't know how to use LaTex but if i NEED to (if the above isn't clear enough) then I will be forced to learn and come back to post it as LaTex.
 
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Lena1992 said:
x^3/1 *(x-2)

is different from

f(x)= x^1/3(x-2)

which is

x^{\frac 1 3}(x-2)

Is that it?

Do you see why we have no idea what the question is?
 


x^{\frac 1 3}(x-2)

Yes it is this. I sincerely apologize for my mistakes.

I'm just trying to derive this using the quotient rule please I need someone to help me!

23mtoa8.jpg
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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