How can you Differentiate X^X

  • Thread starter Noj Werdna
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  • #1
Is there special name for numbers that are to the power of themselves e.g. X^X; 3^3; 4^4
And how can you Differentiate X^X...Thanks

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  • #2
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1. It does have some name I don't remember.
2. In order to differentiate it, rewrite it as [tex]x^{x}=e^{x\ln(x)}[/tex], and use the chain rule.
  • #3
what is it when differentiated? i dont do chain rule yet :(
if you know no need to work it out if not :)
and what is D2y/dx^2?
Thanks...jw :)
  • #4
where's the fun in just knowing the result rather than deriving it yourself?

Also, the symbol you mentioned is the second derivative of y with respect to x. I.e. to find it you take the derivative of y, and then take the derivative of that.
  • #5

[tex]\frac{dy}{dx}[/tex] or f'(x) is the first derivative, which can be differentiated into the second derivative; [tex]\frac{d^2 y}{d x^2}[/tex] or f''(x) if you will.
The third derivative [tex]\frac{d^3 y}{dx^3}[/tex] or f'''(x) is found by differentiating f''(x) i.e., [tex]\frac{d^3 y}{dx^3} = \frac{d}{dx} \left(\frac{d^2 y}{dx^2}\right)[/tex] and so on...

These are called the "higher derivatives".

Now in your question you have to find f'(x) by differentiating [tex]f(x) = x^x[/tex] first, then you have to differentiate the answer again in order to get f''(x).

So, I will give you a hint => we start off by differentiating [tex]x^x[/tex],

[tex]y = x^x[/tex]
[tex]lny = lnx^x[/tex]
[tex]lny = xlnx[/tex]
Differentiate both sides;
[tex]\frac{1}{y} y' = lnx + x \frac{1}{x}[/tex]
So you just have 1/y * y' = lnx + 1
[tex]y' = y(lnx + 1) = x^x (lnx + 1) [/tex]

Now I leave you to differentiate this again in order to obtain [tex]\frac{d^2 y}{d x^2}[/tex].

Hope that helps.
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