Differentiate by first principle $\sqrt{ax+b}$
Let $f\left(x\right)=\sqrt{ax+b}$
$\begin{aligned}
f'\left(x\right) & = \lim\limits_{h \to 0} \; \dfrac{f\left(x+h\right)-f\left(x\right)}{h} \\
& = \lim\limits_{h \to 0} \; \dfrac{\sqrt{a\left(x+h\right)+b}-\sqrt{ax+b}}{h} \\
& = \lim\limits_{h \to 0} \; \dfrac{\left[\sqrt{a\left(x+h\right)+b}-\sqrt{ax+b} \right] \left[\sqrt{a\left(x+h\right)+b}+\sqrt{ax+b}\right]}{h \left[\sqrt{a\left(x+h\right)+b}+\sqrt{ax+b}\right]} \\
& = \lim\limits_{h \to 0} \; \dfrac{a\left(x+h\right)+b-ax-b}{h \left[\sqrt{a\left(x+h\right)+b}+\sqrt{ax+b}\right]} \\
& = \lim\limits_{h \to 0} \; \dfrac{ah}{h \left[\sqrt{a\left(x+h\right)+b}+\sqrt{ax+b}\right]} \\
& = a \; \lim\limits_{h \to 0} \; \dfrac{1}{\sqrt{a\left(x+h\right)+b}+\sqrt{ax+b}} \\
& = \dfrac{a}{\sqrt{ax+b}+\sqrt{ax+b}} \\
& = \dfrac{a}{2\sqrt{ax+b}}
\end{aligned}$