Question

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

Answer 1

The derivative of `ln(x)` is `1/x`. Thus, keeping the constant out of the derivation (it being only a coefficient)...

`(dy)/(dx)=2(1/x)=2/x`

Answer 2

`d/(dx)( lnx) = 1/x`

Using first principles

An expansion on the prior answer

Explanation:

I thought it would be interesting to derive `d/(dx) (ln x)`

We know the definition of the derivative is:

`(d(f(x)))/dx = lim_(h to 0) (f(x+h)-f(x))/h`

`=> d/(dx) ( lnx) = (ln(x+h) - lnx )/h`

Using our log laws:

`lim_(h to 0) (ln( (x+h)/x ) )/h`

We know `(x+h)/x = 1 + h/x`

`=> lim_(h to 0) (ln(1+h/x))/h`

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The not so interesting way:

As `h to 0` the denominator and numirator `-> 0`

So know we can use the L'Hopitals rule:

`=> lim_(h to 0 ) (d/(dh) ( ln(1+h/x) ))/( d/(dh)( h ))`

`=> lim_(h to 0 ) ((1/x)/(1+h/x) )/1`

`= 1/x`

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Or we could use the initial way:

`=> lim_(h to 0) (1+h/x)^(1/h) = e^(1/x)`

As we know:

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`lim_(phi to oo) (1+ 1/(phi))^phi = e`

`lim_(phi to 0) (1+ phi)^(1/phi) = e`

`therefore lim_(phi to 0) (1+ phi/gamma )^(gamma/phi) = e`

Raising each side to `1/gamma` power

`lim_(phi to 0) (1 + phi/gamma)^(1/phi) = e^(1/gamma)`

`color(blue)(--------------)`

Raising each side by the power of `h`:

`=> lim_(h to 0) (1+h/x) = e^(h/x)`

`=> lim_(h to 0) ln(e^(h/x)) /h`

`=> lim_(h to 0) ((h/x)* ln(e))/h`

`( ln(e) = 1 )`

`=> lim_(h to 0) 1/x`

`= 1/x`