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,
$$f'(x)=\lim_{\Delta x\to 0} \dfrac {f(x+\Delta x)-f(x)}{\Delta x}$$
$$=\lim_{\Delta x\to 0} \dfrac {\log (ax+a\Delta x+b)-\log(ax+b)}{\Delta x}$$
$$=\lim_{\Delta x\to 0} \dfrac {\log (\dfrac {ax+a\Delta x+b}{ax+b})}{\Delta x}$$
Best Answer
From where you left off:
$$\lim_{\Delta x\to 0} \dfrac {\log \left(\dfrac {ax+a\Delta x+b}{ax+b}\right)}{\Delta x}$$
$$=\lim_{\Delta x\to 0} \dfrac {\log \left(1 +\dfrac {a\Delta x}{ax+b}\right)}{\Delta x}$$
$$=\lim_{\Delta x\to 0} \dfrac {\log \left(1 +\dfrac {a\Delta x}{ax+b}\right)}{\Delta x} \cdot \frac{a(ax+b)}{a(ax+b)}$$
$$=\lim_{\Delta x\to 0} \dfrac {\log \left(1 +\dfrac {a\Delta x}{ax+b}\right)}{\frac{a\Delta x}{ax+b}} \cdot \frac{a}{ax+b}$$
Since:
$$\lim_{h \to 0} (1+h)^{\frac 1h} = e$$ $$\lim_{h \to 0} \log(1+h)^{\frac 1h} = \log(e)$$ $$\lim_{h \to 0} \frac{\log(1+h)}{h}=1$$
Then: $$\lim_{\Delta x\to 0} \dfrac {\log \left(1 +\dfrac {a\Delta x}{ax+b}\right)}{\frac{a\Delta x}{ax+b}} \cdot \frac{a}{ax+b}$$
$$=1 \cdot \frac{a}{ax+b}$$
$$=\frac{a}{ax+b}$$