What is the derivative of ln(1+x)?

Question

283 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-15T00:56:45

#1/(x+1#

#d/dx [ln (1+x)]#

#=1/(x+1) . d/dx (x+1)#
#=(d/dx (x) + d/dx (1)) / (x+1)#
#=(1+0)/(x+1)#
#1/(x+1)#

Answer 2 · 2018-04-15T04:35:54

#1/(1+x)#

We use the chain rule here, which states that,

#dy/dx=dy/(du)*(du)/dx#

Let #u=1+x,:.du=dx,(du)/dx=1#.

Then #y=lnu,:.dy=1/u \ du,dy/(du)=1/u#.

Combining,

#dy/dx=1/u*1#

#=1/u#

Substitute back #u=1+x# to get:

#=1/(1+x)#