Hi!

I must prove that the function f(x) is continuously differentiable function (just first derivate) on interval (-1,1).

$\displaystyle f(x)=exp(\frac{1}{x^2-1}) \ |x|<1 , \ 0 $ for all other values.

I do that:

$\displaystyle f(x)=e^{\frac{1}{x^2-1}}=e^u \ [u(x)=\frac{1}{x^2-1}]$

$\displaystyle \frac{d}{dx}f(x)=\frac{df}{du}\frac{du}{dx}=-e^u \frac{2x}{(x^2-1)^2}=-\frac{2x}{(x^2-1)^2}e^{\frac{1}{x^2-1}$

Is it ok?

Thanks in advance