Derivative of ln(x) using First Principle of Derivatives

The Mathematician — Sat, 24 Sep 2022 13:00:59 +0000

We will prove the derivative of

\ln(x)

\frac{1}{x}

using the first principle of derivatives. If we have

\log_a(x)

where

a = e

, then

\log_a(x) = \ln(x)

Proof. Let

f(x) = \ln(x)

. Then

\begin{align*} f'(x) &= \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h} \\ &= \lim_{h \rightarrow 0} \frac{\ln(x + h) - \ln(x)}{h} \\ &= \lim_{h \rightarrow 0} \frac{\ln(\frac{x + h}{x})}{h} \\ &= \lim_{h \rightarrow 0} \frac{\ln(1 + \frac{h}{x})}{h} \end{align*}

Now we will multiply the nominator and denominator with

\frac{1}{x}

\begin{align*} \lim_{h \rightarrow 0} \frac{\frac{1}{x}\ln(1 + \frac{h}{x})}{\frac{1}{x}h} &= \frac{1}{x}\lim_{h \rightarrow 0} \frac{x}{h}\ln(1 + h/x) \end{align*}

Now we will use the logarithm power rule

b \cdot \log_a(x) = \log_a(x^b)

\begin{align*} \frac{1}{x}\lim_{h \rightarrow 0} \ln(1 + h/x)^{\frac{x}{h}} = \frac{1}{x} \ln(\lim_{h \rightarrow 0}(1 + h/x)^{\frac{x}{h}}) \end{align*}

We know that

\lim_{h \rightarrow 0} (1 + h/x)^{\frac{x}{h}} = e

\begin{align*} \frac{1}{x} \ln(e) = \frac{1}{x} \cdot 1 = \frac{1}{x}. \end{align*}

So our desired result is

f'(x) = \frac{1}{x}

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