How do you differentiate #y=-lnx#?

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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#