Question #11e5e

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Answer 1 · 2018-02-02T11:11:15

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

let

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

take natural logs of both sides

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

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

differentiating implicitly, and using the product rule on the #RHS#

#d/(dx)(lny)=ln(1+1/x)d/(dx)(x)+xd/(dx)(ln(1+1/x))#

#1/y(dy)/(dx)=ln(1+1/x)+x((d/(dy)(1+1/x))/(1+1/x))#

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

#(dy)/(dx)=y[ln(1+1/x)-1/(x+1)]#

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