Harder problems

1) integrate e^(-1/x^2)  (I couldn't do this one)

   where e is the base of the natural logarithm

2) differentiate x^x (this is easier)

3)White to move, mate in two:

4)White to move, mate in two:

e^(-1/x^2) = what? 0?

The result should be a family of functions (function + integration constant)You should find the antiderivative of e^(-1/x^2) wich I couldn't do and several of my friends couldn't.If anyone has the solution pls post it, I would really like to know it.

I believe, your function f(x) = e^(-1/x^2) cann't integrate (with elementary functions), unless using special functions or series. One online integrator give me this:

F(x) = sqrt(pi)*erf(1/x) + x*e^(-1/x^2) + C

but I don't know, is this right.

Thanks.I tried it with Wolfram Integrator and it gives me the same result.

Symbolic manipulation programs are perfectly reliable, tseta. Why did you doubt it? [Also could check by differentiating]

Elroch, yes they are, if it's known, how to use the program and how it works. In case of online tools, I don't know... So I think, it should take into account - especially, when results doesn't verified. But you are right, perhaps I am too careful with programs.

Ripper89, there are a lot of functions which can't integrate with elementary functions. For example:

• e^(a*x^2): a constant; the normal distribution
• sqrt(1 - (e^2)*(sin x)^2): e constant; the arc length of the ellipse
• (e^x)/x
• sin(a*x + sin(x)): a constant
• and so on...

Hello there

about Your 2 question: x^x = e^(ln x^x) = e^(x lnx). Now it shoud be easy :)

This way it gives me:

d/dx[e^(xlnx)]=(e^xlnx)*d/dx(xlnx)=(e^xlnx)lnx

because d/dx(xlnx)=x'lnx + x (lnx)'=lnx +x*x^(-1)=lnx

However my math teacher told me that the differentiation of x^x should be the combination of the differentiation of a^x and x^a.

So my version is this:

d/dx(x^x)=(x^x)lnx + x*x^(x-1)=(x^x)lnx + x^x =(x^x)(lnx +1)....is this correct?

(x^x)' = [e^(x lnx)]' = e^(x lnx) (x lnx)' = x^x (x lnx)' = x^x (lnx +1)

some ppl use other methods like this:

y = x^x      / ln(...)

ln y = x lnx / (...)'

y'/y = lnx + 1

y' = y(lnx + 1) = x^x (lnx + 1)

Anyway, Your second version its ok. In first (x lx)' shoud be (x' lnx) + [x (lnx)'] = (1 + lnx) + x(1/x) = 1 + lnx

What You answer when teacher will ask: "Is this function it differentiable for all x?" ;)

Sorry, I made a mistake...you are right d/dx(xlnx) is lnx +1

It is not differentiable for all x.The function is defined for all positive real numbers and all negative numbers which are smaller than -1.Between (-1,0) only the odd numbers are ok because the even ones will be even square roots and even square roots of negative numbers are not defined on the real set.

the derivative: the domain of the ln-function is (0, infinity) so the derivative should exist for all positive real numbers except 0.

thats right

Cool