- anonymous

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

- chestercat

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- anonymous

mom I need to attach the equations

- anonymous

##### 1 Attachment

- anonymous

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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## More answers

- anonymous

Your f' is defined only for |R\{-6^{0.5), 6^(0.5)}, actually

- TuringTest

how do you figure @Zekarias

- TuringTest

f' is not defined at +/-6^(0.5), but it is for all other points as far as I see

- TuringTest

I do see one problem which is that for -6^(0.5)<|x|0 the slope should be negative, but the derivative is positive

- TuringTest

-6^(0.5)<|x|<0
I meant

- TuringTest

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)

- anonymous

ok, so how should I take the derivative then? Without rewriting it and using the chainrule over and over?

- TuringTest

that I'm not so sure about. I'm thinking on it.

- TuringTest

@amistre64 any ideas here?

- amistre64

the derivative on the attachment looks fine; im not sure what the question is tho

- TuringTest

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

- TuringTest

but what about the point I brought up?
f' for -6^(1/2)

- amistre64

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

- amistre64

\[u=(x^2-6)^2~:~u'=4x(x^2-6)\]
\[D[u^{1/3}]=\frac{u^{-1/3}}{3}u'\]
\[D[u^{1/3}]=\frac{4x(x^2-6)}{3((x^2-6)^2)^{1/3}}\]

- amistre64

at -1\[\frac{-*-}{+}=+\]

- TuringTest

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

- amistre64

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

- amistre64

the subtlties are in how we are not using the "correct" use of a derivative

- amistre64

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

- anonymous

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?

- amistre64

correct, I would simplify it by making a substitution; then replacing those values in the end

- anonymous

k, thx. I will try that now.

- anonymous

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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