Given the definition of the function: $$ f(x) = e^x - e^{-x} + \frac{2x}{(x^2+1)^2} , x∈(-∞，0]$$
What's the monotonicity of this function in the given range? I tried to calculate the derivation to this function, and I got the result: $$ f'(x) = e^x + e^{-x} + \frac{-6x^4 -4x^2+2}{(x^2+1)^4}$$, the result of the derivation is still a Transcendental Function, and I cannot determine the positive and negative of the derivative function in the range(i.e. x∈(-∞，0]).
How to determine the positive and negative of the derivative function to determine the monotonicity of the f(x) here?
Let's take a look at
$$\frac{-6x^4-4x^2+2}{(x^2+1)^4}$$
We will prove that
$$\frac{-6x^4-4x^2+2}{(x^2+1)^4} > -1$$
This gives us that $$(x^2+1)^4 > 6x^4+4x^2-2$$
$$x^8+4x^6+6x^4+4x^2+1 > 6x^4+4x^2-2$$
$$x^8+4x^6+3 = (x^4)^2 + 4(x^3)^2 + 3 > 0$$
which is true.
Thus, we have shown that the above inequality is true.
And it should be easy to see that $$e^x+e^{-x} \geq 1$$
since $e^{-x} \geq 1$.
Thus, combining the two inequalities we yield $$f'(x) = e^x+e^{-x}+\frac{-6x^4-4x^2+2}{(x^2+1)^4} > 0 = -1+1$$
Thus, the function is monotonically increasing.