  Let $\alpha\in\,]0,1[$. We prove that the existence of the conformable
fractional derivative $T_{\alpha}f$ of a function
$f:[0,\infty[\,\longrightarrow \mathbb{R}$ introduced by Khalil et al. in [R.
Khalil, M. Al Horani, A. Yousef, M. Sababheh, A new definition of fractional
derivative, J. Comput. Appl. Math. 264 (2014) 65-70] is equivalent to classical
differentiability. Precisely the fractional $\alpha$-derivative of $f$ is the
pointwise product $T_{\alpha}f(x)=x^{1-\alpha}f^{\prime}(x)$, $x>0$. This
simplifies the recent results concerning conformable fractional calculus.
