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For function $f(x)$ on $\mathbb{R}$, the directional derivative $f'(x; y)$ on direction $y$ is defined by $f'(x;y)=\lim_{t\to 0+}\frac{f(x+ty)-f(x)}{t}$. If $f$ is differentiable, then $f'(x;y)=yf'(x)$. Now, if $f'_+(x)$ is the right-side derivative at $x$ (not differentiable at $x$, for instance $f(x)=|x|$ at $x=0$), is that true $f'(x;y)=yf_+'(x)$?

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Let $f(x)=-x$ for $x <0$ and $0$ for $x \geq 0$. Then $f'_{+}(0)=0$. take $y=-1$. Then $\frac {f(0+ty)-f(0)} {t}=\frac {-ty} t \to 1$ as $t \to 0+$ whereas $yf'_{+}(0)=0$.

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  • $\begingroup$ Suppose $f(x)$ is additionally symmetric and $f'_+(x) > 0$. Does the assertion hold? $\endgroup$
    – jsmath
    Commented Aug 10, 2019 at 17:13

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