# dy/dx x^1/3

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Find the derivative of [Math]x^{1/3}[/Math] by the first principle.

May anyone show me the steps?

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No, I won't show you step by step, since it is elementary manipulation of variables, let me just remind me of what you need to do

you need to somehow work out:

((x+e)^1/3 - x^1/3)/e

s^3-t^3 = (s-t)(s^2+st+t^2)

now, what happens if you let s^3=x+e and t^3=x?

and put it all together and let e tend to zero... (if done right you should just be able to let e=0 and get the answer.)

edit: too late.

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Find the derivative of [Math]x^{1/3}[/Math] by the first principle.

May anyone show me the steps?

And, there is a thread somewhere that does show it step by step...

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It's exactly the same as how you'd work it out for $x^2$ or any other simple power, except it's using 1/3 and not ^2 or whatever.

Is using this:

$x^n = nx^{n-1}$

considered as using first principle? Or for first principle do you need to use the equation?

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first priciple... (The definition of a derivative)

$f'(x)=\lim_{\Delta x \to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$

so...

$f(x)=x^{1/3}$

$f'(x)=\lim_{\Delta x \to 0}\frac{(x+\Delta x)^{1/3}-x^{1/3}}{\Delta x}$

EDIT:Some people replace "Delta x" with h or y. Matt replaced it with e...

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And some people use a lower case delta, but hey, it's all the same thing.

So if the question (and you were getting marked for calculations) was to derive something via first principle you couldn't use $x^n = nx^{n-1}$ even though it turns out the same answer. Shame that, it's a lot quicker!

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How do you know that your formula holds? You need to prove it when asked to prove it.... Ie, when asked to prove that P is true you cannot simply say since P is true, P is true, which is exactly what you just did.

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So if the question (and you were getting marked for calculations) was to derive something via first principle you couldn't use $x^n = nx^{n-1}$ even though it turns out the same answer. Shame that' date=' it's a lot quicker![/quote']

You're one of those many people who doesn't read math books I take it.

A useful exercise for you would be to go back to your calculus book where that rule is stated. Then continue reading past where it says, "Proof:...", and see how they prove it. It will start from the limit definition of the derivative.

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That was rhetorical right?  referring to matt grime

Well, if not, then sure you have a point, but if someone says what's the derivative of $x^2$ then you can quickly do it in your head just by knowing that

$f(x) = x^2$

so

$f'(x) = 2x^{2-1} = 2x$

You need to prove it when asked to prove it

Obviously just by stating it you have not proved it true. You need to be able to differentiate from first principle but that doesn't mean you can't use $x^n = nx^{n-1}$

 whole post was referring to matt grime, Tom Mattson's post wasn't there when I started typing.

And in response to Tom Mattson's post, well firstly the bit under the quote from matt grime and secondly I learnt differentiation for the 1st time a few days ago!

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Obviously just by stating it you have not proved it true. You need to be able to differentiate from first principle but that doesn't mean you can't use $x^n = nx^{n-1}$

I think you missed my point.

Yes the rule you state is true.

No it is not a "first principle" (as the limit definition of f'(x) is).

No you can't use it when asked to differentiate something from first principles.

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No you can't use it when asked to differentiate something from first principles.

I'm with you now, but I think you missed my point!

The bit in post #7 was not a question, I was stating that you can't use $x^n = nx^{n-1}$ to answer a question which says "using first principle".

Then the bit in post #10 "You need to be able to differentiate from first principle but that doesn't mean you can't use $x^n = nx^{n-1}$" applies for finding the derivative... it implies the question does NOT say "using first principle".

Like I said, I learnt all this a few days ago!

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I think you'll find it was the bit in post 5 where you don't appear to know what "from first principles" means that is causing the head scratching by me (and I'd guess Tom). You see you appear to have initally thought it was ok to use the formula, but now you're saying you agree it isn't acceptable.

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And then after my question in post5 BobbyJoeCool replied in post6... which to me cleared the little "well you get the same answer so can you use it?" debate in my head, which was why I said in post7 as a statement that if the question said using first principle you couldn't use the other method.

[/confusion] !

 In reply to CanadaAotS's post below (editing this because it's not worth another post... I GET IT!!! I understood with certainty in post #7, forget I ever mentioned it!!!

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look, first principle means you must use this equation:

$f'(x) = \lim_{h\to0}\frac{f(x+h) - f(x)}{h}$

so you can't use any of the shortcut methods you were talking about.

(however you could use them to get the answer so you know what to look for when using the 'first principle')

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So for $f(x) = x^1/3$ you'd have to find the derivative using

$f'(x) = \lim_{h\to0} \frac{f[(x + h)^{1/3}] - f(x^{1/3})}{h}$

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Which has already been said in post #6.... and then you solve from there.

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little nitpick... you say:

So for $f(x) = x^1/3$ you'd have to find the derivative using

$f'(x) = \lim_{h\to0} \frac{f[(x + h)^{1/3}] - f(x^{1/3})}{h}$

but if f(x)=cubed root x' date=' and you say

[math']f'(x) = \lim_{h\to0} \frac{f[(x + h)^{1/3}] - f(x^{1/3})}{h}[/math]

What you're really saying is...

$f'(x) = \lim_{h\to0} \frac{[(x + h)^{1/3}]^{1/3} - (x^{1/3})^{1/3}}{h}$

because you already put the (x+h) into the funciton when you put the (1/3) as the exponent, but now you're saying (by taking f((x+h)^(1/3))) that you want to put it into the function a second time...

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$f(x) = x^{1/3}$

$f'(x) = \lim_{h\to0}\frac{(x+h)^{1/3} - x^{1/3}}{h}$

$f'(x) = \lim_{h\to0}\frac{(x+h)^{\frac{1}{3}} - x^{\frac{1}{3}}}{h}\left(\frac{(x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}(x)^{\frac{1}{3}} + x^{\frac{2}{3}}}{(x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}(x)^{\frac{1}{3}} + x^{\frac{2}{3}}}\right)$

$f'(x) = \lim_{h\to0}\frac{x + h - x}{h\left((x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}(x)^{\frac{1}{3}} + x^{\frac{2}{3}\right)}$

$f'(x) = \lim_{h\to0}\frac{h}{h\left((x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}(x)^{\frac{1}{3}} + x^{\frac{2}{3}\right)}$

$f'(x) = \lim_{h\to0}\frac{1}{\left((x+h)^{\frac{2}{3}} + (x+h)^{\frac{1}{3}}(x)^{\frac{1}{3}} + x^{\frac{2}{3}\right)}$

$f'(x) = \lim_{h\to0}\frac{1}{\left((x+h)^{\frac{2}{3}} + (x^{2}+hx)^{\frac{1}{3}} + x^{\frac{2}{3}\right)}$

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No, I won't show you step by step, since it is elementary manipulation of variables, let me just remind me of what you need to do

you need to somehow work out:

((x+e)^1/3 - x^1/3)/e

s^3-t^3 = (s-t)(s^2+st+t^2)

now, what happens if you let s^3=x+e and t^3=x?

and put it all together and let e tend to zero... (if done right you should just be able to let e=0 and get the answer.)

Thanks

Is using this:

x^n = nx^{n-1}

considered as using first principle? Or for first principle do you need to use the equation?

This is not considered as using the first principle.

The approach in post 6 is

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How come I can't edit post 19? By the way. Follow that step in post 19, the last visible one I mean. It becomes a difference of perfect cubes on the top, aka x+h - x. that gets you pretty much where you need to be to solve the problem.

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How come I can't edit post 19?

You can only edit a post up to 4 hours after you've posted it. It's a rule we have on here to stop people from deliberately removing all of their forum posts, which breaks the continuity of the thread.

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