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.
what are the steps and explanations for differantiating this function?
y = ln(e^-x + xe^-x)
3 answers
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....
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....
ln of e^a= a by definition.
What does log mean if not that?
What does log mean if not that?