Is the derivative always nonnegative in a neighbourhood of a minimum?

While you can probably come up with some explicit combination of exponential and trigonometric functions that is a counterexample, I find it much more enlightening to instead just cobble one together with bump functions.

Start with a smooth function $\varphi:[0,1]\to\mathbb{R}$ which is identically $0$ in neighborhoods of $0$ and $1$, nonnegative on $[0,1/2]$, nonpositive on $[1/2,1]$ (and negative somewhere) and has positive integral. (So, it jumps up to positive values somewhere in the middle of $[0,1/2]$, jumps down to negative values somewhere in $[1/2,1]$, and the positive values have a larger integral than the negative values.)

Now pick a shrinking sequence of disjoint intervals $[a_n,b_n]$ approaching $0$ and consider a function $g:[0,1]\to\mathbb{R}$ which is $0$ except on the intervals $[a_n,b_n]$, and on each $[a_n,b_n]$ is given by $g(x)=c_n\varphi(\frac{x-a_n}{b_n-a_n})$ for some $c_n>0$. If we pick the coefficients $c_n$ to shrink fast enough, then all the derivatives $g^{(k)}(x)$ will approach $0$ as $x\to 0$ and so $g$ will be smooth even at $0$.

Finally, define $f(x)=\int_0^xg(t)\,dt$. Then $f$ is smooth because $g$ is. Also, $f(x)>0=f(0)$ for all $x>0$, by our choice of $\varphi$, since the integral of $\varphi$ is positive and moreover the integral of $\varphi$ over $[0,s]$ is still nonnegative for any $s\in [0,1]$ (so if $x$ is in the middle of one of the intervals $[a_n,b_n]$, the integral of $g$ over the first part of that interval will not be negative). But $f'=g$ is negative on points of every interval $[a_n,b_n]$, and these points get arbitrarily close to $0$.