For $s=\sigma+it$, $t\geq 8$, $1-\frac{1}{2}(\log t)^{-1}\leq \sigma\leq 2$, \begin{equation} \zeta(s)\ll \log t \text{ and } \zeta'(s)\ll (\log t)^2 \end{equation}
The first part can easily been proved, even for $1-\log(t)^{-1}\leq \sigma\leq 3$, but I am not quite sure about the second part. The authos claims that the estimation for $\zeta'$ immediately follows from the Cauchy formula \begin{equation} \zeta'(s)=\frac{1}{2\pi i}\int_{\vert z-s\vert=r} \frac{\zeta(z)}{(z-s)^2}\mathrm{dz} \end{equation} with $r=\frac{1}{2}(\log t)^{-1}$. And it took my quite long to figure out why, but I am not sure if my proof is correct: I start with
\begin{align*} \zeta'(s)&=\frac{1}{2\pi i}\int_{\vert z-s\vert=r} \frac{\zeta(z)}{(z-s)^2}\mathrm{dz}\\ &=\frac{1}{2\pi i} \int_0^{2\pi}\frac{\zeta(s+re^{i\varphi})}{(s+re^{i\varphi}-s)^2}ire^{i\varphi}\mathrm{d\varphi}\\ &=\frac{1}{2\pi i}\int_0^{2\pi}\zeta(s+re^{i\varphi})\frac{i}{r}e^{-i\varphi}\mathrm{d\varphi} \end{align*}
From this point I see that $1-\frac{1}{\log t}\leq\mathrm{Re}(s+\frac{e^{i\varphi}}{2\log t})\leq 3$, so we can apply our first estimation for $\zeta(s)$ that gives us (we also get rid of constants) \begin{align*} &\frac{1}{2\pi i}\int_0^{2\pi}\zeta(s+re^{i\varphi})\frac{i}{r}e^{-i\varphi}\mathrm{d\varphi}\\ &\ll \log t\int_0^{2\pi}\bigg\vert \frac{1}{(2\log t)^{-1}}e^{-i\varphi}\bigg\vert\mathrm{d\varphi}\\ &=(\log t)^2\int_0^{2\pi}\bigg\vert e^{-i\varphi}\bigg\vert\mathrm{d\varphi} = (\log t)^2 \end{align*}
I don't know how to describe it but this proof feels kinda weird. Can anyone review this proof if it is legit to argue like I do? Am I missing any requirements about the asymptotic behavier?