Question

what are the steps and explanations for differantiating this function?

y = ln(e^-x + xe^-x)

Answer 1

replace the <> function with u.
y=ln u
y'=1/u du/dx

where du/dx is d/dx (e^-x(1+ x))
to that as a d(vw)/dx

Then put it all together.

Answer 2

thanks, but I have another question.

How does ln(e^-x) = -x

I know that the derivative of ln is 1/x so it would be 1/(e^-x) but how does that simplify to -x? I am confused....

Answer 3

ln of e^a= a by definition.

What does log mean if not that?