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,
$$f'(x)=\lim_{\Delta x\to 0} \dfrac {f(x+\Delta x)-f(x)}{\Delta x}$$
$$=\lim_{\Delta x\to 0} \dfrac {\log (ax+a\Delta x+b)-\log(ax+b)}{\Delta x}$$
$$=\lim_{\Delta x\to 0} \dfrac {\log (\dfrac {ax+a\Delta x+b}{ax+b})}{\Delta x}$$
 A: 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:
$$\lim_{\Delta x\to 0} \dfrac {\log \left(1 +\dfrac {a\Delta x}{ax+b}\right)}{\frac{a\Delta x}{ax+b}} \cdot \frac{a}{ax+b}$$
$$=1 \cdot \frac{a}{ax+b}$$
$$=\frac{a}{ax+b}$$
A: You are almost done! Now, just recall: $$\lim_{x \to 0} \frac{\log(1+x)}{x} = 1$$
Using this, we get: $$\lim_{\delta x \to 0} \frac{\log(1+\frac{a\Delta x}{ax+b})}{\Delta x} = \frac{\log(1+\color{red}{\frac{a\Delta x}{ax+b})}}{\color{red}{\frac{a\Delta x}{ax+b}}} \times \frac{a}{ax+b} = \frac{a}{ax+b}$$
A: Assuming the limit $(1+t)^{1/t} \to e$ as $t \to 0$ is known we have
$$(1+at)^{1/t} = \left( (1+at)^{1/(at)} \right)^a \to e^a$$
Therefore,
$$
\dfrac {\log (\dfrac {ax+a\Delta x+b}{ax+b})}{\Delta x}
= \dfrac {\log \left(1+\dfrac {a}{ax+b}\Delta x\right)}{\Delta x} 
= \log \left(1+\dfrac {a}{ax+b}\Delta x\right)^\frac{1}{\Delta x} \\
\to \log e^{\frac {a}{ax+b}}
= \frac {a}{ax+b}
$$
