Why does $xy^x$ decrease for low $y$?

Question

I'm trying to understand the behaviour of $f\left( x,y\right) =xy^{x}.$ I've computed its derivative with respect to $x$ and got

$$\frac{\partial f\left( x,y\right) }{\partial x}=y^{x}+xy^{x}\ln y=y^{x}\left( 1+\ln y^{x}\right) . $$ Given $x_{0}>0,$ the derivative is negative for all $y\in ]0,\exp{-\frac{1}{x_{0}}}[ $ and positive if $y\in ]\exp{-\frac{1}{x_{0}}},+\infty[ .$ What I'd like to have is some intuition as to what causes $f\left( x,y\right) $ to decrease in the first place (I think I see why it increases).

Answer 1

It would decrease when $1+\ln(y^x)=1+x \ln y<0,$ or when $\ln(y)<-1/x,$ assuming $x>0.$

I think this agrees with what you found.

Answer 2

For $0<y<1$, $y^x$ is a decreasing function of $x$. When $y$ is very small, this decreasing is more rapid than the increase of $x$. For some $0 < \eta(x) < 1$, when $y$ surpasses $\eta$, the decaying of $y^x$ slows to less than the increase of $x$, so $f$ begins to increase. As $x$ gets large, $y^x$ will dominate $x$ in terms of asymptotic behavior, so we expect $\eta \to 1$ as $x\to\infty$, so that $f$ mirrors the behavior of $y^x$: if $0 < y< 1$, $f\to 0$, but if $y$ surpasses $1$, $f\to \infty$. On the edge case where $y=1$, $y^x \equiv 1$, so the behavior of $f$ mirrors that of $x$.