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

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Answer 1 · 2017-02-05T15:02:49

#f'(x) = (1+x)^(1/x)(1/(x(x+1)) -ln(1+x)/x^2)#

Consider:

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

The function is defined only for #(1+x) > 0# so we can take its logarithm:

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

and using the properties of logarithms:

#ln(f(x)) = 1/xln( 1+x)#

Now differentiate both sides, using the chain rule at first member and the product rule at the second member:

# (f'(x))/f(x) = 1/(x(x+1)) -ln(1+x)/x^2#

and since we know #f(x)#:

#f'(x) = (1+x)^(1/x)(1/(x(x+1)) -ln(1+x)/x^2)#

Alternatively we can write the function as:

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

using the chain rule:

#d/dx e^(ln(1+x)/x) = e^(ln(1+x)/x) d/dxln(1+x)/x = (1+x)^(1/x)d/dxln(1+x)/x #

which is the same expression we had in the other way.