Question

Prove using the first principals that `d/dx(lnx) =1/x` ?

Answer 1

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

Explanation:

We know that ,

`color(red)((1)lnX-lnY=ln(X/Y),where, X and Y inRR^+`

`color(blue)((2)n*lnX=ln(X^n) ,where , X inRR^+ and n in RR`

Let ,

`f(x)=lnx=>f(x+h)=ln(x+h)`

Differentiating using First principles :

`color(green)(f'(x)=lim_(hto0)(f(x+h)-f(x))/h)`

`=>f'(x)=lim_(hto0)(ln(x+h)-lnx)/h`

`=>f'(x)=lim_(hto0)1/h{color(red)(ln(x+h)-lnx)}`

`=>f'(x)=lim_(hto0)1/h{color(red)(ln((x+h)/x))}tocolor(red)(Apply (1)`

`=>f'(x)=lim_(hto0)1/hln(1+h/x)`

`=>f'(x)=lim_(hto0)1/x*color(blue)(x/hln(1+h/x)`

`=>f'(x)=1/x*lim_(hto0)color(blue)(ln(1+h/x)^(x/h)toApply(2)`

`=>f'(x)=1/x*ln[lim_(hto0)(1+h/x)^(x/h)]`

Now, `color(brown)(hto0 =>h/xto0 and h/x=n=>lim_(nto0)(1+n)^(1/n)=e`

So,

`f'(x)=1/x*ln[color(brown)(lim_(h/xto0)(1+h/x)^(x/h))]`

`=>f'(x)=1/x*ln[color(brown)(e)] ,where, lne=1`

`=>f'(x)=1/x`

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