calculus - Find, from first principle, the derivative of:

Find, from first principle, the derivative of:

$$\log (ax+b)$$

My Attempt:

$$f(x)=\log (ax+b)$$ $$f(x+\Delta x)=\log (ax+a\Delta x+b)$$ Now, $$f'(x)=\lim_{\Delta x\to 0} \dfrac {f(x+\Delta x)-f(x)}{\Delta x}$$ $$=\lim_{\Delta x\to 0} \dfrac {\log (ax+a\Delta x+b)-\log(ax+b)}{\Delta x}$$ $$=\lim_{\Delta x\to 0} \dfrac {\log (\dfrac {ax+a\Delta x+b}{ax+b})}{\Delta x}$$