Question

How do you differentiate `y=-lnx`?

Answer 1

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

Explanation:

If `f(x)=axxg(x)`, `(df)/(dx)=axx(dg)/(dx)`

Hence as `y=-lnx=-1xxlnx`

`(dy)/(dx)=-1xx1/x=-1/x`

For Derivative of `lnx` see below

`d/(dx) lnx=Lt_(h->0)(ln(x+h)-lnx)/h`

= `Lt_(h->0)1/hln((x+h)/x)`

= `Lt_(h->0)ln(1+h/x)^(1/h)` - assuming `u=h/x`

= `Lt_(h->0)ln(1+u)^(1/ux)`

= `Lt_(u->0)ln((1+u)^(1/u))^(1/x)`

= `1/xLt_(u->0)ln(1+u)^(1/u)`

= `1/x xx lne`

= `1/x xx1=1/x`