How do you differentiate #y=(1+x)^(1/x)#?

Question

1923 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-31T20:53:26

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

We have:

# y = (1+x)^(1/x) #

First note that the function is not defined when #x=0#. We can take (natural) logarithms of both sides to get:

# lny = ln{(1+x)^(1/x)} #
# " " = 1/x \ ln(1+x) #
# " " = ln(1+x)/x #

Differentiating implicitly and applying the quotient rule and the chain rule gives:

# 1/y dy/dx = { (x)(1/(1+x)) - (ln(1+x))(1) } / (x)^2 #
# " " = { x/(1+x) - ln(1+x) } / (x)^2 #
# " " = 1/(x(1+x)) - ln(1+x)/x^2 #

And so:

# dy/dx = y {1/(x(1+x)) - ln(1+x)/x^2} #
# " " = (1+x)^(1/x) {1/(x(1+x)) - ln(1+x)/x^2} #

Answer 2 · 2017-03-31T20:54:14

The answer is #=(1+x)^(1/x)(1/(x(1+x))-ln(1+x)/x^2)#

We diiferentiate using logs.

#y=(1+x)^(1/x)#

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

#lny=1/x*ln(1+x)#

Differentiating

#(lny)'=(1/x*ln(1+x))'#

#1/y*dy/dx=1/x*1/(1+x)-1/x^2*ln(1+x)#

#dy/dx=y(1/(x(1+x))-1/x^2ln(1+x))#

#dy/dx=(1+x)^(1/x)(1/(x(1+x))-ln(1+x)/x^2)#