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reyler

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The squaring function ƒ(x) = x² is differentiable at x = 3, and its derivative there is 6. This result is established by writing the difference quotient as follows:

 

 

 

Ex:  {f(3+h)-f(3)\ h} = {(3+h)^2 - 9\over{h}} = {9 + 6h + h^2 - 9\{h}} = {6h + h^2\{h}} = 6 + h.

 

Oh yes, I'm not a Pharaoh :)

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I'm impressed but there's a couple of things I don't agree with. Please correct me if I'm wrong.

This is what you wrote:

{f(3+h)-f(3)\ h} = {(3+h)^2 - 9\over{h}} = {9 + 6h + h^2 - 9\{h}} = {6h + h^2\{h}} = 6 + h.

Doesn't: 6h + h^2\{h} = 6h + h; which factorises into = h(6+1) = 7h 

 

Also: From {[glow=red,2,300]9[/glow] + 6h + h^2 - 9\{h}}  to {6h + h^2\{h}}, I want to know where the 9 disappears to.

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Vampires are generally good at everything, although I do better with languages.

 

The last expression shows that the difference quotient equals 6 + h when h is not zero and is undefined when h is zero. (Remember that because of the definition of the difference quotient, the difference quotient is never defined when h is zero.) However, there is a natural way of filling in a value for the difference quotient at zero, namely 6. Hence the slope of the graph of the squaring function at the point (3, 9) is 6, and so its derivative at x = 3 is ƒ '(3) = 6.

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Hoo, yeah "I'm on a Boat" by Lonely Island is daaaaa best. i really wanna quote but pg13 tells me otherwise but still i think it is a pg13 song anyway. have you heard 13 year olds these days??

Also, "like a boss" both my favourites.

have you also heard?: ahhmm :-[

-**** in a box

-who said we're wack

-**** in my pants

 

lonely island is coolio

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First what is arcsin?

 

Do you know the trig ratio sine? Well when graphing it, the equation looks like this: [y = sin x]. But create the inverse of this and you get: [x = sin y]. But this is difficult to work with so we re-arrange again to get [y = arcsin x]. You see [x = sin y] = [y = arcsin x].

 

It's differentiable is:

          1            __

(SQRT[1 - x^2] )

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