In order to show that a function is C^1 is it enough to show that the 1. partial derivatives exists?

Question

Hello I am having some issues with the following exercise:

Let $\textbf{h}: \mathbb{R}^2 \rightarrow \mathbb{R}$ given by

$\large \textbf{h}(u,v) = u^2 + (v-1)^2 - 5 + e^{u-2}$

(i) Show that $\textbf{h}(2,1) = 0$ and $\textbf{h}\in C^1(\mathbb{R}^2)$

$\large \textbf{h}(2,1) = 2^2 + (1-1)^2 - 5 + e^{2-2} = 4+0-5+1 = 0$

and next I show that $\textbf{h}\in C^1(\mathbb{R}^2)$. I start out by showing that the partial derivatives exists:

\begin{align*} \large \textbf{h}_u(u, v)=& \lim_{k\to 0}\frac{\textbf{h}(u+k, v)- \textbf{h}(u,v)}{k}\\ =& \lim_{k\to 0}\frac{\left((u+k)^2 + (v-1)^2 - 5 + e^{(u+k)-2}\right)- \left((u)^2 + (v-1)^2 - 5 + e^{(u-2)}\right)}{k}\\ =& \lim_{k\to 0}\frac{(u+k)^2 + (v-1)^2 - 5 + e^{(u+k)-2} - (u)^2 - (v-1)^2 + 5 - e^{(u-2)}}{k}\\ =& \lim_{k\to 0}\frac{(u+k)^2 + e^{(u+k)-2} - (u)^2 - e^{(u-2)}}{k}\\ =& \lim_{k\to 0}\left(2(u+k) + e^{(u+k)-2}\right)\\ =& \lim_{k\to 0}\left(2u+2k + e^{(u+k)-2}\right)\\ =& 2u + 0 + e^{(u+0)-2}\\ = & 2u + e^{(u-2)} \end{align*}

\begin{align*} \large \textbf{h}_v(u, v)=& \lim_{k\to 0}\frac{\textbf{h}(u, v+k)- \textbf{h}(u,v)}{k}\\ =& \lim_{k\to 0}\frac{\left(u^2 + ((v+k)-1)^2 - 5 + e^{(u-2)}\right)- \left((u)^2 + (v-1)^2 - 5 + e^{(u-2)}\right)}{k}\\ =& \lim_{k\to 0}\frac{((v+k)-1)^2 - (v-1)^2}{k}\\ =& \lim_{k\to 0}\frac{(v+k)^2+1-2(v+k) - v^2-1+2v}{k}\\ =& \lim_{k\to 0}\frac{v^2+k^2 + 2vk+1-2v-2k - v^2-1+2v}{k}\\ =& \lim_{k\to 0}\frac{k^2 + 2vk-2k}{k}\\ =& \lim_{k\to 0}\left(2k + 2v-2\right) \\ =&2(v-1) \end{align*}

And now since I have shown that the partial derivatives exists I can conclude that $h(u,v) \in C^1(\mathbb{R}^2)$. Is this correct and enough?

Answer 1

Your $h$ (not ${\bf h}$, since $h$ is scalar-valued) is $C^\infty$ "by inspection".

The existence of the partial derivatives $h_u(u_0,v_0)$, $\>h_v(u_0,v_0)$ is necessary, but not sufficient, for the differentiability of $h$ at $(u_0,v_0)$. Examples like $$h(u,v):={uv\over u^2+v^2}\quad\bigl((u,v)\ne(0,0)\bigr),\qquad h(0,0):=0$$ have appeared many times on MSE.

The following fact, however, takes care of $99\%$ of the cases: If the partial derivatives $h_u$, $h_v$ exist and are continuous on the open set $\Omega\subset{\mathbb R}^2$ then $h$ is continuously differentiable, in short: $C^1$, on $\Omega$.

Answer 2

Hint: I think that the map $h$ is composite of three $C^{\infty}$ function and by chain rule the map $h$ is $C^{\infty}$.