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Derivative of the logarithm function

Let $\log_a(x)$ be the logarithm function to base $a$. We will prove that: $\log_{a}'(x)=\frac{\log_a{e}}{x}$.

Prerequisite information[edit]

The proof[edit]

From the definition of the derivative:

[math]\begin{eqnarray} \log_{a}'(x) & = & \lim_{h\to 0} \frac{\log_a(x+h)-\log_a(x)}{h}\end{eqnarray}[/math]

Then we use the properties of the logarithm function:

[math]\begin{eqnarray}& = & \lim_{h \to 0}\frac{1}{h} \log_a\left(\frac{x+h}{x}\right) \\ & = & \lim_{h \to 0}\log_a\left[\left(\frac{x+h}{x}\right)^{\frac{1}{h}}\right] \\ & = & \log_a\left[\lim_{h \to 0}\left(\frac{x+h}{x}\right)^{\frac{1}{h}}\right] \\ & = & \log_a\left[\lim_{h \to 0}\left(1+\frac{h}{x}\right)^{\frac{1}{h}}\right] \\ & = & \log_a\left[\left[\lim_{\frac{x}{h} \to \infty}\left(1+\frac{1}{\frac{x}{h}}\right)^{\frac{x}{h}}\right]^{\frac{1}{x}} \right] \end{eqnarray}[/math]

From the definition of the Euler's number [math]\lim_{n \to + \infty}\left(1+\frac{1}{n}\right)^n=e[/math]:

[math]\begin{eqnarray} &=& \log_a\left(e^{\frac{1}{x}}\right) \\ &=& \frac{\log_a(e)}{x}\#\end{eqnarray}[/math]

Corollary[edit]

A well-known corollary of Derivative of the logarithm function|the derivative of the logarithm function is:

[math]\log_{a}'(x)=\frac{1}{x\ln(a)}[/math]

The proof[edit]

We first change the base of the logarithm in order to get the natural logarithm:

[math]\begin{eqnarray} \log_{a}(e)\frac{1}{x} & = & \frac{\ln(e)}{\ln(a)}\frac{1}{x} \end{eqnarray}[/math]

From the defnition of the natural logarithm [math]\ln(e)=1[/math], we have:

[math]\begin{eqnarray} & = & \frac{1}{x\ln(a)} \end{eqnarray}[/math]

Note[edit]

It is important to note that it is possible to deduce the derivative of the natural logarithm from this formula:

[math]\ln'(x)=\frac{1}{x}[/math]