Find, from first principle, the derivative of:

Question

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

What definition do you take for $log(x)$? It is usually defined as $\int_{1}^{x} \frac{1}{t} dt$, which would mean the result is straightforward. — Higurashi, Dec 23 '17 at 13:28
the result is given by $$\frac{a}{ax+b}$$ if you compute the limit — Dr. Sonnhard Graubner, Dec 23 '17 at 13:32
@CaveJohnson I think first principle means the limit definition of derivative (if exists). This terminology was used in my secondary school. — Alex Vong, Dec 23 '17 at 16:13

score 9 · Answer 1 · answered Dec 23 '17 at 13:36

9

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: $$\lim_{\Delta x\to 0} \dfrac {\log \left(1 +\dfrac {a\Delta x}{ax+b}\right)}{\frac{a\Delta x}{ax+b}} \cdot \frac{a}{ax+b}$$

$$=1 \cdot \frac{a}{ax+b}$$

$$=\frac{a}{ax+b}$$

answered Dec 23 '17 at 13:36

Josh

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I’m confused: isn’t $\frac{d}{dx}log(ax+b)=\frac{a}{(ax+b)*ln 10}$ – Daniel Dec 23 '17 at 16:03
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Since the base was not explicitly stated, I am assuming that it is the logarithm to the base of e, however, the answer you provided is correct if the logarithm is instead to the base of 10. – Josh Dec 23 '17 at 16:12
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A good rule of thumb for when someone says 'logarithm' is: Mathematicians mean base $e$, computer folk mean base 2, and engineers mean base 10. – JonathanZ Dec 23 '17 at 16:16

score 4 · Answer 2 · answered Dec 23 '17 at 13:37

You are almost done! Now, just recall: $$\lim_{x \to 0} \frac{\log(1+x)}{x} = 1$$

Using this, we get: $$\lim_{\delta x \to 0} \frac{\log(1+\frac{a\Delta x}{ax+b})}{\Delta x} = \frac{\log(1+\color{red}{\frac{a\Delta x}{ax+b})}}{\color{red}{\frac{a\Delta x}{ax+b}}} \times \frac{a}{ax+b} = \frac{a}{ax+b}$$

score 2 · Answer 3 · answered Dec 23 '17 at 13:38

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, $$ \dfrac {\log (\dfrac {ax+a\Delta x+b}{ax+b})}{\Delta x} = \dfrac {\log \left(1+\dfrac {a}{ax+b}\Delta x\right)}{\Delta x} = \log \left(1+\dfrac {a}{ax+b}\Delta x\right)^\frac{1}{\Delta x} \\ \to \log e^{\frac {a}{ax+b}} = \frac {a}{ax+b} $$

Find, from first principle, the derivative of:

3 Answers3