How do you differentiate #y=1/lnx#?

Question

45214 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-11T18:06:10

#=- 1/(x (ln x)^{2} )#

you can do this simply as #( (ln x)^{-1})'#

#=- (ln x)^{-2} (ln x)'#

#=- (ln x)^{-2} 1/x#

#=- 1/(x (ln x)^{2} )#

if you want to fiddle about with e and logs i suppose you could say that

#1/y = ln x#

#e^(1/y) = e^ln x = x#

so
#(e^(1/y))' = 1#

and
#( e^(1/y))' = e^(1/y) (1/y)'#

#= e^(1/y) * -(1/y^2) y'#

So #- e^(1/y) (1/y^2) y' = 1#

# y' = -y^2 * 1 / e^(1/y)#

# = -(1/ln x)^2 * 1/x#

#=- 1/(x (ln x)^{2} )#

same but bit more involved and fiddly