Monotonicity of functions Let $f(x) = xe^{x^2} + e^{-x^2}$
I'd like to prove that this function increases monotonically in the interval $(0,1)$.
I was able to do it by taking the derivative and proving it was greater than zero.
(though it took me quite some time to do that)
I wanted to know if there was any other more elegant method to do the same.
 A: The derivative is easy to compute, really, assuming I didn't screw up! Using an obvious abbreviation, it's
$$(1+2x^2)e^+-2xe^- > (1-2x+2x^2)e^+ =\left (2\left(x-\frac 1 2\right)^2 + \frac 1 2\right)e^+ > 0$$
A: $$g:(0,1)\rightarrow (1,e), x\mapsto e^x, \forall x,y\in(0,1), x<y \Rightarrow g(x)<g(y)$$
$$h:(0,1)\rightarrow (0,1), x\mapsto x^2, \forall x,y\in(0,1), x<y \Rightarrow h(x)<h(y)$$
$$g\circ h:(0,1)\rightarrow (1,e), x\mapsto e^{x^2}, \forall x,y\in(0,1), x<y \Rightarrow g(h(x))<g(h(y))$$
$$j:(1,e)\rightarrow \left(1,{1\over e}\right), x\mapsto {1\over x}, \forall x,y\in(0,1), x<y \Rightarrow j(x)>j(y)$$
$$j\circ g \circ h:(0,1)\rightarrow \left(1,{1\over e}\right), x\mapsto {1\over e^{x^2}}, \forall x,y\in(0,1), x<y \Rightarrow j(g(h(x)))>j(g(h(y)))$$
Let $u(x)$ be a monotonically increasing function
$$p:(0,1)\rightarrow (0,e), x\mapsto x \cdot u(x), \forall x,y\in(0,1), x<y \Rightarrow p(x)<p(y)$$
With $u(x)=g \circ h$ 
$$p:(0,1)\rightarrow (0,e), x\mapsto x \cdot g(h(x)),$$ That is $$p:(0,1)\rightarrow (0,e), x\mapsto xe^{x^2}, \forall x,y\in(0,1), x<y \Rightarrow p(x)<p(y)$$
As $g\circ h$ increase at the same speed than $j\circ g \circ h$ decrease, 
$xe^{x^2}$ is increasing faster than $e^{-x^2}$ decrease.
Conclusion : $f(x)=xe^{x^2}+e^{-x^2}$ is increasing monotically.
