# How do you find the derivative of ln x?

Apr 18, 2015

$\frac{1}{x}$

Consider f(x) = lnx and using the first principles work out $f ' \left(x\right) = \lim h \to 0 \frac{f \left(x + h\right) - f \left(x\right)}{h}$

= $\lim h \to 0 \frac{\ln \left(x + h\right) - \ln x}{h}$

=$\lim h \to 0 \frac{\ln \left(\frac{x + h}{x}\right)}{h}$

= $\lim h \to 0 \frac{\ln \left(1 + \frac{h}{x}\right)}{h}$

= $\lim h \to 0 \frac{1}{x} \frac{\ln \left(1 + \frac{h}{x}\right)}{\frac{h}{x}}$

= $\frac{1}{x}$ $\lim h \to 0 \frac{\ln \left(1 + \frac{h}{x}\right)}{\frac{h}{x}}$

= $\frac{1}{x}$, because $\left[\lim h \to 0 \frac{\ln \left(1 + \frac{h}{x}\right)}{\frac{h}{x}} = 1\right]$

Apr 18, 2015

The "how" depends on the definition you are using for $\ln x$.

I like: $\ln x = {\int}_{1}^{x} \frac{1}{t} \mathrm{dt}$ 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:

$y = \ln x$ $\iff$ $x = {e}^{y}$

Differentiate implicitely:

$\frac{d}{\mathrm{dx}} \left(x\right) = \frac{d}{\mathrm{dx}} \left({e}^{y}\right)$

$1 = {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}}$

So
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{e} ^ y = \frac{1}{x}$