Question

Find the derivative using first principles? : `ln(3x)`

Answer 1

`dy/dx = 1/x`

Explanation:

Let `f(x)=y=ln(3x)`

Then the limit definition of the derivative tells us that;

`f'(x) = lim_(h rarr 0) ( f(x+h)-f(x) ) / h`
`\ \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( ln(3(x+h))-ln(3x) ) / h`
`\ \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( ln(3x+3h)-ln(3x) ) / h`
`\ \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( ln( (3x+3h)/(3x) ) ) / h`
`\ \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( ln( 1+h/x ) ) / h`

Now perform a substitution, Let:

`v=h/x` so that `v rarr 0` as `h rarr 0`

Then:

`f'(x) = lim_(v rarr 0) ( ln( 1+v ) ) / (vx)`
`\ \ \ \ \ \ \ \ \ = lim_(v rarr 0) 1/x( ln( 1+v ) ) / (v)`
`\ \ \ \ \ \ \ \ \ = 1/x \ lim_(v rarr 0) (1/v) ln( 1+v )`
`\ \ \ \ \ \ \ \ \ = 1/x \ lim_(v rarr 0) ln( 1+v ) ^(1/v)`
`\ \ \ \ \ \ \ \ \ = 1/x \ ln{lim_(v rarr 0) ( 1+v ) ^(1/v)}`

This limit is a known result, as determined by Leonhard Euler, and we have:

`lim_(v rarr 0) ln( 1+v ) ^(1/v) = e`

Leading to the result:

`f'(x) = 1/x \ lne`
`\ \ \ \ \ \ \ \ \ = 1/x`