How do you find the derivative of #(1+x)^(1/x)#?

1 Answer
Shaikh A.
Dec 5, 2016

#dy/dx=(1+x)^(1/x)/x(1/(1+x)-ln(1+x)/x)#

Explanation:

#y=(1+x)^(1/x)# Taking ln both sides
#lny=1/xln(1+x)#
differentiate both sides
#1/ydy/dx= 1/x(1/(1+x))+ln(1+x)(-1/x^2)#
#dy/dx=[1/x(1/(1+x))+ln(1+x)(-1/x^2)]y#
#dy/dx=1/x[1/(1+x)-ln(1+x)/x)]y#

By putting value of y in R.H.S. to make R.H.S. in term of x
#dy/dx=1/x[1/(1+x)-ln(1+x)/x)][(1+x)^(1/x)#

#dy/dx=(1+x)^(1/x)/x(1/(1+x)-ln(1+x)/x)#

Related questions
See all questions in Chain Rule
Impact of this question
18524 views around the world
You can reuse this answer
Creative Commons License