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

Question

1886 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-16T12:45:41

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

The function:

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

has only positive values, so we can take its logarithm:

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

Differentiate now the equation above:

#d/dx ln (f(x)) = d/dx (ln(1+x)/x)#

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

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

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