Can you calculate a partial derivative wrt x if the y coordinate isn't fixed?

Question

Specifically, I would like to know if $u:\mathbb{R}^2\to\mathbb{R}$ is continuous at $(x_0,y_0)$ and has continuous partial derivatives of first order at $(x_0,y_0)$, does

$$\lim_{h\to 0}\frac{u(x_0+h,y_0+h)-u(x_0,y_0+h)}h=u_x(x_0,y_0)$$

Where $u_x$ is the partial of $u$ w.r.t. x?

This is actually coming from a complex analysis problem in which I am trying to show that if $f=u+iv$, and each of $u$ and $v$ are continuous and have continuous partial derivatives at some $z\in\mathbb{C}$, and if the limit of the modulus of the difference quotient exists, then either $f$ or $\overline{f}$ are differentiable at $z$. If I can prove the aforementioned result, I can prove this.

I have tried thinking of $u$ as a function of $h$ and using the mean value theorem to simplify this as is common when dealing with derivatives of multi-variable functions, but the $h$ in the second coordinate complicates this, and I haven't found a way around it.

Any help with this or tips are greatly appreciated!

Answer 1

We write \begin{align} u(x_0+h,y_0+h)-u(x_0,y_0+h)&=\left(u(x_0,y_0+h)+h\partial_{x}u(x_0,y_0+h)+O(h^2)\right)-u(x_0,y_0+h)\\ &=h\partial_{x}u(x_0,y_0+h)+O(h^2) \end{align} so that \begin{align} \frac{1}{h}\left(u(x_0+h,y_0+h)-u(x_0,y_0+h)\right)&=\partial_{x}u(x_0,y_0+h)+O(h)\to\partial_{x}u(x_0,y_0) \end{align} as $h\to0$, by continuity of the partial derivative $\partial_{x}u$ at the point $(x_0,y_0)$.