What is the derivative of this function: $\frac {d}{dx}x^{\lfloor{x}\rfloor}?$

Question

What is the derivative of the following function?
$$\frac {d}{dx}x^{\lfloor{x}\rfloor}$$ Here, $\lfloor x \rfloor$ is the floor function.

I tried: $$\frac {d}{dx} x^x=\frac {d}{dx} e^{x \ln x}=x^x (\ln x +1)+C$$

But, here $\lfloor{x}\rfloor$ is problematic for me.

Answer 1

Suppose $x\in[n,n+1)$, where $n$ is an integer. Then $$ f(x)=x^n $$ and so $f$ coincides with $x^n$ over the open interval $(n,n+1)$. Hence $f'(x)=nx^{n-1}$ for $x\in(n,n+1)$.

Thus, for noninteger $x$, $$ f'(x)=\lfloor x\rfloor x^{\lfloor x\rfloor-1} $$ The problem is now to see whether $f$ is differentiable at integers. But, if $n\ne0$ is an integer $$ \lim_{x\to n^+}x^{\lfloor x\rfloor}=\lim_{x\to n^+}x^n=n^n $$ whereas $$ \lim_{x\to n^-}x^{\lfloor x\rfloor}=\lim_{x\to n^-}x^{n-1}=n^{n-1} $$ The limits are different when $n\ne1$. Also $$ \lim_{x\to0^+}f(x)=1 \qquad \lim_{x\to0^-}f(x)=-\infty $$ This shows the function is not continuous at the integers $\ne1$, so not differentiable either.

Now try and see if the derivative exists at $x=1$.

Answer 2

The function has discontinuities at integer values of $x$. Other than that, you can evaluate the derivatives using $\frac{d}{dx}x^a=ax^{a-1}$ because $\lfloor x\rfloor$ is constant for $a < x < a+1$, where $a \in \mathbb{Z}$. To summarize, the derivative of $x^{\lfloor x \rfloor}$ when it exists is $$\lfloor x \rfloor x^{\lfloor x \rfloor-1}.$$