Question

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

Answer 1

mom I need to attach the equations

Answer 2

My f`is apparently only true for x>0 and I have no clue where I make that assumption in my differentiation.

Answer 3

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

Answer 4

how do you figure @Zekarias

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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)<x<0, but the derivative would be positive as you have it

Answer 9

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

Answer 10

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

Answer 11

@amistre64 any ideas here?

Answer 12

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

Answer 13

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

Answer 14

but what about the point I brought up?
f' for -6^(1/2)<x<0 should be negative, but it's positive

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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?

Answer 23

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

Answer 24

k, thx. I will try that now.

Answer 25

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?