How can you Differentiate X^X

[Noj Werdna]

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

 

[arildno]

1. It does have some name I don't remember.
2. In order to differentiate it, rewrite it as $$x^{x}=e^{x\ln(x)}$$, and use the chain rule.

 

[Noj Werdna]

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 :)

 

[uman]

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.

 

[roam]

Hello!!!

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

These are called the "higher derivatives".

Now in your question you have to find f'(x) by differentiating $$f(x) = x^x$$ 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 $$x^x$$,

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

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

Hope that helps.