# Differentiability $\implies$ partial derivatives exist?

Say I have a differentiable mapping $$f: G \to \Bbb{R}$$, where $$G$$ is an open subset of $$\Bbb{R^n}$$. This implies that the directional derivative $$\frac{\partial f}{\partial u}(x_0)$$ exists in every direction ( except in the direction $$u=0$$ ). Now, since the partial derivative is a special case of the directional derivative, may we conclude that the partial derivatives $$\frac{\partial f}{\partial x_i}(x_0)$$, where $$x_0 = (x_1, ... x_n )$$, also exist?

• Yes. All partial derivatives are directional derivatives in some coordinates, and differentiability is maintained under linear maps. (Note I'm awkwardly trying to use what you already know here. Proving the result directly is just as easy) Sep 23 '20 at 13:17
• I personally think the directional derivative in the direction $u=0$ should just be $0$. There is no part of the definition of directional derivative that actually requires $u\neq0$ to make sense, so why should one limit the definition? Sep 23 '20 at 13:17

For sure the partial derivatives $$\frac{\partial f}{\partial x_i}(x_0)$$ do exist and if $$L = f^\prime(x_0)$$ is the derivative of $$f$$ at $$x_0$$, we have

$$\frac{\partial f}{\partial x_i}(x_0) = L(e_i)$$ where $$e_i$$ is the $$i$$-th vector of the canonical basis.