## TomLikesPhysics 3 years ago I can only find derivative for x>0 , but I don't know why (according to wolfram).

1. TomLikesPhysics

mom I need to attach the equations

2. TomLikesPhysics

3. TomLikesPhysics

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

4. Zekarias

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

5. TuringTest

how do you figure @Zekarias

6. TuringTest

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

7. 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

8. TuringTest

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

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

10. TomLikesPhysics

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

11. TuringTest

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

12. TuringTest

@amistre64 any ideas here?

13. amistre64

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

14. 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

15. TuringTest

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

16. 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

17. amistre64

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

18. amistre64

at -1$\frac{-*-}{+}=+$

19. 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

20. 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

21. amistre64

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

22. 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

23. TomLikesPhysics

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?

24. amistre64

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

25. TomLikesPhysics

k, thx. I will try that now.

26. TomLikesPhysics

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?