Find the first principal derivative of logx

April 29, 2026Mathematics

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Answer

\frac{1}{x}

5 steps

The first principal derivative of $\log x$ is found using the limit definition of the derivative:

\frac{d}{dx}(\log x) = \lim_{h \to 0} \frac{\log(x+h) - \log x}{h}

Use the logarithm property $\log a - \log b = \log \frac{a}{b}$ .
$\lim_{h \to 0} \frac{\log\left(\frac{x+h}{x}\right)}{h}$
Rewrite the expression inside the logarithm.
$\lim_{h \to 0} \frac{\log\left(1 + \frac{h}{x}\right)}{h}$
Rearrange the expression to use the limit $\lim_{u \to 0} \frac{\log(1+u)}{u} = 1$ .
Let $u = \frac{h}{x}$ . As $h \to 0$ , $u \to 0$ . We need to manipulate the expression to match this form.
$\lim_{h \to 0} \frac{1}{x} \cdot \frac{\log\left(1 + \frac{h}{x}\right)}{\frac{h}{x}}$
Apply the limit.
$\frac{1}{x} \cdot \lim_{u \to 0} \frac{\log(1+u)}{u}$ $\frac{1}{x} \cdot 1$
Simplify.
$\frac{1}{x}$
The first principal derivative of $\log x$ is $\boxed{\frac{1}{x}}$ .

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