Gabriela Gallegos

2023-03-06

How to find the derivative of ln x?

Tommy Kerr

Consider f(x) = lnx and using the first principles work out
$f\prime \left(x\right)=limh\to 0\frac{f\left(x+h\right)-f\left(x\right)}{h}$
= $limh\to 0\frac{\mathrm{ln}\left(x+h\right)-\mathrm{ln}x}{h}$
=$limh\to 0\frac{\mathrm{ln}\left(\frac{x+h}{x}\right)}{h}$
= $limh\to 0\frac{\mathrm{ln}\left(1+\frac{h}{x}\right)}{h}$
= $limh\to 0\frac{1}{x}\frac{\mathrm{ln}\left(1+\frac{h}{x}\right)}{\frac{h}{x}}$
= $\frac{1}{x}$ $limh\to 0\frac{\mathrm{ln}\left(1+\frac{h}{x}\right)}{\frac{h}{x}}$
= $\frac{1}{x}$, because $\left[limh\to 0\frac{\mathrm{ln}\left(1+\frac{h}{x}\right)}{\frac{h}{x}}=1\right]$

kancsalrma

The "how" depends on the definition you are using for ln x.
$\mathrm{ln}x={\int }_{1}^{x}\frac{1}{t}dt$ and the result is immediate useing the Fundamental Theorem of Calculus.
If you start with a definition of ${e}^{x}$ and find its derivative, then you probably defined ln x by:
$y=\mathrm{ln}x$ $⇔$ $x={e}^{y}$
Differentiate implicitely:
$\frac{d}{dx}\left(x\right)=\frac{d}{dx}\left({e}^{y}\right)$
$1={e}^{y}\frac{dy}{dx}$
Hence, $\frac{dy}{dx}=\frac{1}{{e}^{y}}=\frac{1}{x}$

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