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I can only find derivative for x>0 , but I don't know why (according to wolfram).

Mathematics
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mom I need to attach the equations
1 Attachment
My f`is apparently only true for x>0 and I have no clue where I make that assumption in my differentiation.

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Other answers:

Your f' is defined only for |R\{-6^{0.5), 6^(0.5)}, actually
how do you figure @Zekarias
f' is not defined at +/-6^(0.5), but it is for all other points as far as I see
I do see one problem which is that for -6^(0.5)<|x|0 the slope should be negative, but the derivative is positive
-6^(0.5)<|x|<0 I meant
http://www.wolframalpha.com/input/?i=plot%20y%3D(x%5E2-6)%5E(2%2F3)&t=crmtb01 you can see the slope is negative for -6^(1/2)
ok, so how should I take the derivative then? Without rewriting it and using the chainrule over and over?
that I'm not so sure about. I'm thinking on it.
@amistre64 any ideas here?
the derivative on the attachment looks fine; im not sure what the question is tho
http://www.wolframalpha.com/input/?i=derivative+of+y%3D%28x%5E2-6%29%5E%282%2F3%29 true I don't see wolf giving the condition that x>0
but what about the point I brought up? f' for -6^(1/2)
http://www.wolframalpha.com/input/?i=derivative+cbrt%28%28x%5E2-6%29%5E2%29 i think this has more intricate workings than we think
\[u=(x^2-6)^2~:~u'=4x(x^2-6)\] \[D[u^{1/3}]=\frac{u^{-1/3}}{3}u'\] \[D[u^{1/3}]=\frac{4x(x^2-6)}{3((x^2-6)^2)^{1/3}}\]
at -1\[\frac{-*-}{+}=+\]
but that should not be if you look at the graph http://www.wolframalpha.com/input/?i=plot%20y%3D(x%5E2-6)%5E(2%2F3)&t=crmtb01 should be f'<0 at x=-1
notice that the function\[\sqrt[3]{((x^2-6)^2)}\ne \left(\sqrt[3]{(x^2-6)}\right)^2\]at all points http://www.wolframalpha.com/input/?i=%28%28x%5E2-6%29%5E2%29%5E%281%2F3%29+-+%28%28x%5E2-6%29%5E%281%2F3%29%29%5E2
the subtlties are in how we are not using the "correct" use of a derivative
the way we are used to working with exponents seems to be a misuse of notation and doesnt express the full nature of the problem http://www.wolframalpha.com/input/?i=y%3D%28%28x%5E2-6%29%5E2%29%5E%281%2F3%29%2C+y%3D+%28%28x%5E2-6%29%5E%281%2F3%29%29%5E2
So I should not rewrite the equation and just use the chainrule multiple times then to end up with the same answer as wolfram does?
correct, I would simplify it by making a substitution; then replacing those values in the end
k, thx. I will try that now.
Damn. That went well and it was pretty quick and easy too. So the first attempt did not work because I messed the exponents up?

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