How do you find the derivative of #y=e^(-2x-1) * ln(-2x-1)#?

1 Answer
Gav
Jun 21, 2018

#-e^(-2x-1)/(-2x-1)-2ln(-2x-1)e^(-2x-1)#

Explanation:

#d/dxln(-2x-1)e^(-2x-1)#
=#d/dx[ln(-2x-1)]*e^(-2x-1)+ln(-2x-1)*d/dx(e^[-2x-1])#
=#1/(-2x-1)*d/dx(-2x-1)*e^[-2x-1]+e^[-2x-1]*d/dx[-2x-1]*ln(-2x-1)#
=#{(-2*d/x[x]+d/dx[-1])e^[-2x-1]}/[-2x-1]+(-2*d/x[x]+d/dx[-1])ln(-2x-1)e^(-2x-1)#
=#[(0-2*1)e^(-2x-1)]/[-2x-1]+(0-2*1)ln(-2x-1)e^(-2x-1)#
=#-e^(-2x-1)/(-2x-1)-2ln(-2x-1)e^(-2x-1)#

Related questions
See all questions in Differentiating Exponential Functions with Base e
Impact of this question
3240 views around the world
You can reuse this answer
Creative Commons License