# Find the derivative of ln x from first principles

`(d[lnx])/(dx) = lim_(h-gt0)(ln(x+h)-ln(x))/h`

`(d[lnx])/(dx) = lim_(h-gt0)(ln((x+h)/x))/h`

`(d[lnx])/(dx) = lim_(h-gt0)(ln(1+h/x))/h`

Let `v = h/x.` Then, `h-gt0` gives us `v-gt0`

Changing the limits,

`(d[lnx])/(dx) = lim_(v-gt0)(ln(1+v))/(vx)`

`(d[lnx])/(dx) =1/x lim_(v-gt0)(ln(1+v))/v`

`(d[lnx])/(dx) =1/x lim_(v-gt0)1/vln(1+v)`

`(d[lnx])/(dx) =1/x lim_(v-gt0)ln(1+v)^(1/v)`

We can take the limit inside the ln function.

`(d[lnx])/(dx) =1/x ln[lim_(v-gt0)(1+v)^(1/v)]`

Let `u = 1/v` ,...

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`(d[lnx])/(dx) = lim_(h-gt0)(ln(x+h)-ln(x))/h`

`(d[lnx])/(dx) = lim_(h-gt0)(ln((x+h)/x))/h`

`(d[lnx])/(dx) = lim_(h-gt0)(ln(1+h/x))/h`

Let `v = h/x.` Then, `h-gt0` gives us `v-gt0`

Changing the limits,

`(d[lnx])/(dx) = lim_(v-gt0)(ln(1+v))/(vx)`

`(d[lnx])/(dx) =1/x lim_(v-gt0)(ln(1+v))/v`

`(d[lnx])/(dx) =1/x lim_(v-gt0)1/vln(1+v)`

`(d[lnx])/(dx) =1/x lim_(v-gt0)ln(1+v)^(1/v)`

We can take the limit inside the ln function.

`(d[lnx])/(dx) =1/x ln[lim_(v-gt0)(1+v)^(1/v)]`

Let `u = 1/v` , then, when `v -gt0 , u -gt oo`

changing the limits again,

`(d[lnx])/(dx) =1/x ln[lim_(u-gtoo)(1+1/u)^u]`

By definition of e, `lim_(u-gtoo)(1+1/u)^u = e`

Therefore,

`(d[lnx])/(dx) =1/x ln[e]`

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

Proved.