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, $$f'(x)=\lim_{\Delta x\to 0} \dfrac {f(x+\Delta x)-f(x)}{\Delta x}$$ $$=\lim_{\Delta x\to 0} \dfrac {\log (ax+a\Delta x+b)-\log(ax+b)}{\Delta x}$$ $$=\lim_{\Delta x\to 0} \dfrac {\log (\dfrac {ax+a\Delta x+b}{ax+b})}{\Delta x}$$
$\endgroup$ 53 Answers
$\begingroup$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: $$\lim_{\Delta x\to 0} \dfrac {\log \left(1 +\dfrac {a\Delta x}{ax+b}\right)}{\frac{a\Delta x}{ax+b}} \cdot \frac{a}{ax+b}$$
$$=1 \cdot \frac{a}{ax+b}$$
$$=\frac{a}{ax+b}$$
$\endgroup$ 3 $\begingroup$You are almost done! Now, just recall: $$\lim_{x \to 0} \frac{\log(1+x)}{x} = 1$$
Using this, we get: $$\lim_{\delta x \to 0} \frac{\log(1+\frac{a\Delta x}{ax+b})}{\Delta x} = \frac{\log(1+\color{red}{\frac{a\Delta x}{ax+b})}}{\color{red}{\frac{a\Delta x}{ax+b}}} \times \frac{a}{ax+b} = \frac{a}{ax+b}$$
$\endgroup$ $\begingroup$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, $$ \dfrac {\log (\dfrac {ax+a\Delta x+b}{ax+b})}{\Delta x} = \dfrac {\log \left(1+\dfrac {a}{ax+b}\Delta x\right)}{\Delta x} = \log \left(1+\dfrac {a}{ax+b}\Delta x\right)^\frac{1}{\Delta x} \\ \to \log e^{\frac {a}{ax+b}} = \frac {a}{ax+b} $$
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