Question

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

Answer 1

`-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)`