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

2 Answers
Jan 31, 2016

Answer:

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

Explanation:

#y=(x+1)^x#

lets use #ln# in the two sides of the equation,

#lny=xln(x+1)#

now lets differentiate the two sides in respect of #x#

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

#or,1/yd/dx=x*1/(x+1)d/dx(x+1)+lnx#

#or,1/y(dy)/(dx)=x/(x+1)+lnx#

#or,(dy)/(dx)=y(x/(x+1)+lnx)#

#or,(dy)/(dx)=(x+1)^x(x/(x+1)+lnx)#

Jan 31, 2016

Answer:

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

Explanation:

I think the easiest way to do these kinds of problems (with a variable power) without memorizing the formula is through implicit differentiation.

#y=(x+1)^x#

Take the natural logarithm of both sides.

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

Rewrite using logarithm rules.

#ln(y)=xln(x+1)#

Differentiate both sides with respect to #x#.

The left side will spit out a #dy/dx# term, thanks to the chain rule. The right side will require the product rule.

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

Note that differentiating #ln(x+1)# requires chain rule as well, but since #d/dx(x+1)=1# it doesn't make a visible difference.

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

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

Now, solve for #dy/dx# by multiplying both sides by #y#. However, remember that #y=(x+1)^x#.

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