What is the derivative of #y=(1+x)^(1/x)#?

Question

6184 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-01T14:16:00

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

#=[(1+x)^(1/x-1){x-(1+x)ln(1+x)}]/x^2#

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

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

#:. lny=1/xln(1+x)=(ln(1+x))/x#

#:. d/dxlny=d/dx{(ln(1+x))/x}#

#:. d/dylny*dy/dx={xd/dxln(1+x)-ln(1+x)d/dxx}/x^2#

Here, we have used the Chain Rule & the Quotient Rule.

#:. 1/ydy/dx={x/(1+x)-ln(1+x)}/x^2={x-(1+x)ln(1+x)}/(x^2(1+x)#

#:. dy/dx=[y{x-(1+x)ln(1+x)}]/(x^2(1+x)#

Using #(star)#, we get,

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

#=[(1+x)^(1/x-1){x-(1+x)ln(1+x)}]/x^2#