How do you find the derivative of #x+2^x#?

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Answer 1 · 2017-09-11T09:12:09

#1+2^xln2#

because differentiation is a linear operator we have:

#d/(dx)(x+2^x)=d/(dx)(x)+d/(dx)(2^X)#
#=1+d/(dx)(2^x)---(1)#

let us do the second differentiation separately.

let #y=2^x#

#=>lny=ln(2^x)=xln2#

#d/(dx)(lny)=d/(dx)(xln2)#

#1/ydy/(dx)=ln2#

#dy/(dx)=yln2=2^xln2#

#:." "(1)" becomes "1+2^xln2#