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

1 Answer

Oct 10, 2015

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

Explanation:

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

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

#lny=1/xln(1+x)#

#d/dx(lny)=d/dx(1/xln(1+x))#

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

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

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