An apparent counterexample of the derivative proprety of Fourier transform

I learned that one of the properties of the Fourier transform is that $\mathcal{F}[\frac{df}{dx}] = ik\mathcal{F}[f(x)]$. However it seems to me that the function $f(x) = 1$ is a counterexample of this property. Its Fourier transform is $2\pi\delta(k)$ but its deravitive is $0$ and the Fourier transform of $0$ is just $0$. So why it isn't?

There is no contradiction: on the left we have for $f=1$ $$ \mathcal F(f') = \mathcal F(\text{the zero function})=\text{the zero function}$$ and on the right we have $$ ik\mathcal F(f)=i2\pi k \delta(k) = \text{the zero distribution}$$ The result holds for all tempered distributions $f\in\mathcal S'$. Proof follows, though it won't be intelligible if you don't know what a distribution is. Let $\phi$ be a Schwartz function. Then: \begin{align} (\mathcal F(f'),\phi) = (f',\mathcal F\phi) = -(f,(\mathcal F\phi)')=(f,\mathcal F( ik\phi)) = (\mathcal Ff,ik\phi) =(ik\mathcal Ff,\phi),\ \text{QED.} \end{align} And also, the proof that $k\delta(k) = 0$: Let $\phi$ be Schwartz; then, $$ (k\delta(k), \phi(k)) = (\delta(k), k\phi(k))= k\phi(k)\Big|_{k=0} = 0,\ \text{QED.}$$