# continuous partial derivative implies total differentiable (check)

Let $f: \mathbb{R^n}\rightarrow \mathbb{R}$ to have continuous partial derivatives, it suffice to show that $f$ is total differentiable at $(0,..,0)$ with $f_{x_i}(0,..,0) = 0$. Since each partial derivative is continuous, we can rewrite $f$ as $$f(0,..,0) + \int_0^{x_1} f_{x_1}(t_1,0,..,0) dt_1 + \int_0^{x_2} f_{x_2}(x_1,t_2,0,..,0) dt_2 + ... +\int_0^{x_n} f_{x_n}(x_1,x_2,.., t_n) dt_n,$$

and to show $$\lim_{x\rightarrow 0; x\in\mathbb{R}^n/\{0\}} \frac{f(x) - f(0) - h\cdot\triangledown f(0)}{|x|} = 0,$$ by assumption $\triangledown f(0) \equiv 0$, it suffice to show $$\lim_{x\rightarrow 0} \frac{1}{|x|} \left[\int_0^{x_1} f(t_1,0,..,0) dt_1 + \int_0^{x_2} f(x_1,t_2,0,..,0) dt_2 + ... +\int_0^{x_n} f(x_1,x_2,.., t_n) dt_n\right] = 0.$$

For $0<|x|\leq r$, $$\frac{1}{|x|}\left[\int_0^{x_1} f(t_1,0,..,0) dt_1 + \int_0^{x_2} f(x_1,t_2,0,..,0) dt_2 + ... +\int_0^{x_n} f(x_1,x_2,.., t_n) dt_n\right]\\ \leq\frac{n \max_i|x_i|}{|x|} \max_i \left[\sup_{0<|x|\leq r} |f_{x_i} (x) |\right].$$

And as $r\rightarrow 0$, the above goes to zero as well since $f_{x_i}(0) = 0$ and continuous.

If partial derivatives exist in a neighborhood of $\vec 0$ and have finite limits at $\vec 0$, then $f$ is differentiable at $\vec 0$.
Indeed, subtracting a linear function from $f$, we can ensure this finite limit is zero. Let $B$ be a ball centered at $\vec 0$ on which all partial derivatives are bounded by $\epsilon$. The mean value theorem, applied in the direction of $x_i$ (with other variables fixed) implies $$|f(x_1,\dots,x_i,\dots,x_n)-f(x_1,\dots,x_i',\dots,x_n)|\le \epsilon |x_i-x_i'|$$ Applying this $n$ times and summing, we obtain $$|f(x_1,\dots,x_n)-f(\vec 0)|\le \epsilon \sum_i |x_i|$$ Thus $f$ is differentiable at $\vec 0$, with zero gradient.