Derivative of log x base a using First Principle of Derivatives

The Mathematician — Wed, 21 Sep 2022 13:00:52 +0000

We will prove the derivative of

\log_a(x)

using the first principle of derivatives. We will show that the derivative of

\log_a(x)

\frac{1}{x\ln(a)}

, and we will prove that by the definition of derivative.

Proof. Let

f(x) = \log_a(x)

. Then

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

But we are not done. The logarithm base change rule tells us that

\log_a(x) = \frac{\log_b(x)}{\log_b(a)}

. In our case, we will take

b = e

. So we get

\begin{align*} \frac{1}{x} \log_a(e) &= \frac{1}{x} \frac{\log_e(e)}{\log_e(a)} \\ &= \frac{1}{x} \frac{1}{\ln(a)} \\ &= \frac{1}{x \ln(a)} \end{align*}

because

\log_e = \ln

. So our desired result is

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

log base a of x Archives - Epsilonify