I am looking for a 1 continuous variable even bounded function. At the moment, I can only thing in $\tanh x$, $\sin x, \cos x$ and $\frac{1}{(1+x^2n)}$. Can anyone list me some other apart from the previous ones?
Also, is there anyone which its derivative is also even?
On
You can't list all bounded continuous even functions. What is your need? You can also combine even functions to even functions, i.e. $f(x)=\frac{\tanh(x)}{1+\cos(x)+\sin(|x|)}$ and so on. You can also use $\frac1{1+|x|}$.
But $\sin$ is an odd function since $\sin(-x)=-\sin(x)$.
And if $f$ is not constant, even and differentiable, you get $f(x)=f(-x)$. Now derivate both sides using chain rule on the RHS and you get $f'(x)=-f'(-x)$. Therefore $f'$ has to be odd.
FIRST QUESTION: There are infinitely many bounded even continuous functions. Furthermore, if you have an even function $f(x)$ and any other function $g(x)$, the function $$g(f(x))$$ will also be even. This allows you to generate as many as you like.
Furthermore, the sum, difference, product, and ratio of two even functions is also even. Or you can take it even farther. If $g(x_1,...,x_n)$ is some function and $f_1(x),...,f_n(x)$ are all even functions, then $$g(f_1(x),...,f_n(x))$$ is even as well.
SECOND QUESTION: The only function that is even whose derivative is also even is a constant function. This is because if $f(x)$ is even, then $$f(x)=f(-x)$$ and so, by differentiating both sides with respect to $x$, $$f'(x)=-f'(-x)$$ and so $f'(x)$ can only be even if $f'(x)=-f'(x)$, or when $f'(x)=0$, or when $f(x)=C$, where $C$ is a constant. Otherwise, its derivative will always be odd, not even.