# What is the derivative of 2ln(x)?

May 23, 2015

The derivative of $\ln \left(x\right)$ is $\frac{1}{x}$. Thus, keeping the constant out of the derivation (it being only a coefficient)...

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \left(\frac{1}{x}\right) = \frac{2}{x}$

Dec 15, 2017

$\frac{d}{\mathrm{dx}} \left(\ln x\right) = \frac{1}{x}$

Using first principles

An expansion on the prior answer

#### Explanation:

I thought it would be interesting to derive $\frac{d}{\mathrm{dx}} \left(\ln x\right)$

We know the definition of the derivative is:

$\frac{d \left(f \left(x\right)\right)}{\mathrm{dx}} = {\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

$\implies \frac{d}{\mathrm{dx}} \left(\ln x\right) = \frac{\ln \left(x + h\right) - \ln x}{h}$

Using our log laws:

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

We know $\frac{x + h}{x} = 1 + \frac{h}{x}$

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

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The not so interesting way:

As $h \to 0$ the denominator and numirator $\to 0$

So know we can use the L'Hopitals rule:

$\implies {\lim}_{h \to 0} \frac{\frac{d}{\mathrm{dh}} \left(\ln \left(1 + \frac{h}{x}\right)\right)}{\frac{d}{\mathrm{dh}} \left(h\right)}$

$\implies {\lim}_{h \to 0} \frac{\frac{\frac{1}{x}}{1 + \frac{h}{x}}}{1}$

$= \frac{1}{x}$

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Or we could use the initial way:

$\implies {\lim}_{h \to 0} {\left(1 + \frac{h}{x}\right)}^{\frac{1}{h}} = {e}^{\frac{1}{x}}$

As we know:

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${\lim}_{\phi \to \infty} {\left(1 + \frac{1}{\phi}\right)}^{\phi} = e$

${\lim}_{\phi \to 0} {\left(1 + \phi\right)}^{\frac{1}{\phi}} = e$

$\therefore {\lim}_{\phi \to 0} {\left(1 + \frac{\phi}{\gamma}\right)}^{\frac{\gamma}{\phi}} = e$

Raising each side to $\frac{1}{\gamma}$ power

${\lim}_{\phi \to 0} {\left(1 + \frac{\phi}{\gamma}\right)}^{\frac{1}{\phi}} = {e}^{\frac{1}{\gamma}}$

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Raising each side by the power of $h$:

$\implies {\lim}_{h \to 0} \left(1 + \frac{h}{x}\right) = {e}^{\frac{h}{x}}$

$\implies {\lim}_{h \to 0} \ln \frac{{e}^{\frac{h}{x}}}{h}$

$\implies {\lim}_{h \to 0} \frac{\left(\frac{h}{x}\right) \cdot \ln \left(e\right)}{h}$

$\left(\ln \left(e\right) = 1\right)$

$\implies {\lim}_{h \to 0} \frac{1}{x}$

$= \frac{1}{x}$