How do you differentiate $y=-lnx$ ?

Question

1500 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-28T11:08:00

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

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 $u=h/x$

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

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

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

= $1/x xx lne$

= $1/x xx1=1/x$

How do you differentiate y=-lnxy=-lnx?