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

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TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.0mom I need to attach the equations

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.0My f`is apparently only true for x>0 and I have no clue where I make that assumption in my differentiation.

Zekarias
 3 years ago
Best ResponseYou've already chosen the best response.0Your f' is defined only for R\{6^{0.5), 6^(0.5)}, actually

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0how do you figure @Zekarias

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0f' is not defined at +/6^(0.5), but it is for all other points as far as I see

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0I do see one problem which is that for 6^(0.5)<x0 the slope should be negative, but the derivative is positive

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.06^(0.5)<x<0 I meant

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=plot%20y%3D(x%5E26)%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

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.0ok, so how should I take the derivative then? Without rewriting it and using the chainrule over and over?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0that I'm not so sure about. I'm thinking on it.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0@amistre64 any ideas here?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2the derivative on the attachment looks fine; im not sure what the question is tho

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=derivative+of+y%3D%28x%5E26%29%5E%282%2F3%29 true I don't see wolf giving the condition that x>0

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0but what about the point I brought up? f' for 6^(1/2)<x<0 should be negative, but it's positive

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2http://www.wolframalpha.com/input/?i=derivative+cbrt%28%28x%5E26%29%5E2%29 i think this has more intricate workings than we think

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2\[u=(x^26)^2~:~u'=4x(x^26)\] \[D[u^{1/3}]=\frac{u^{1/3}}{3}u'\] \[D[u^{1/3}]=\frac{4x(x^26)}{3((x^26)^2)^{1/3}}\]

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2at 1\[\frac{*}{+}=+\]

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0but that should not be if you look at the graph http://www.wolframalpha.com/input/?i=plot%20y%3D(x%5E26)%5E(2%2F3)&t=crmtb01 should be f'<0 at x=1

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2notice that the function\[\sqrt[3]{((x^26)^2)}\ne \left(\sqrt[3]{(x^26)}\right)^2\]at all points http://www.wolframalpha.com/input/?i=%28%28x%5E26%29%5E2%29%5E%281%2F3%29++%28%28x%5E26%29%5E%281%2F3%29%29%5E2

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2the subtlties are in how we are not using the "correct" use of a derivative

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2the 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%5E26%29%5E2%29%5E%281%2F3%29%2C+y%3D+%28%28x%5E26%29%5E%281%2F3%29%29%5E2

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.0So 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
 3 years ago
Best ResponseYou've already chosen the best response.2correct, I would simplify it by making a substitution; then replacing those values in the end

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.0k, thx. I will try that now.

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.0Damn. 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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