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

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Answer 1 · 2015-12-20T06:00:22

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

#y=(x+1)^x#

Take the natural log of both sides.

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

Simplify using logarithm rules.

#lny=xln(x+1)#

Take the derivative of both sides.

#(y')/y=ln(x+1)+x/(x+1)#

Multiply both sides by #y#, which equals #(x+1)^x#.

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