Let $f,f_n:D(0,1)\to\mathbb{C}$ be holomorphic. Suppose $f_n$ converges uniformly on compacta to $f$. Show that $f'_n$ converges uniformly on compacta to $f'$.
I started with trying manipulating $|f'_n(z)-f'(z)|$ with a boundary $|f(z)|\leq M$ for all $z$ in a compact set $S\subset D(0,1)$, but it doesn't seem to lead to anything. I would like for some help, thanks.
Fix $z_0\in D(0,1)$ and $r>0$ such that $\lvert z-z_0\rvert<r\implies z\in D(0,1)$. Then, if $\lvert z-z_0\rvert<r$,$$f(z)=\frac1{2\pi i}\int_{\lvert z-z_0\rvert=r}\frac{f(\eta)}{\eta-z}\,\mathrm d\eta.$$Since the set $\{z\in\mathbb C\,|\,\lvert z-z_0\vert=r\}$ is compact, it follows from this that $(f_n)_{n\in\mathbb N}$ converges uniformly to $f$ on $D(z_0,r)$.
Now, if $K\subset D(0,1)$ is compact, all you have to do is to consider, for each $z_0\in K$ a disk $D(z_0,r)\subset K$. Since $K$ is compact, it can be covered by finitely many such disks and now you can use the fact that the convergence is uniform on each such disk to prove that the convergence is uniform on $K$.
In order to prove that $(f_n')_{n\in\mathbb N}$ converges uniformly to $f'$ on each compact, you use the same idea, together with the formula$$f'(z)=\frac1{2\pi i}\int_{\lvert z-z_0\rvert=r}\frac{f(\eta)}{(\eta-z)^2}\,\mathrm d\eta.$$