Logarithmic differentiation: y=ln(e^-x + xe^-x)?

Answer 1

For one thing, this isn't really logarithmic differentiation (in the conventional sense), as logarithmic differentiation is usually used to solve questions like this:

Find dy/dx if y = x^[sin(x)]

But with that aside, if

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

First, let's factor e^(-x) within the ln.

y = ln( e^(-x) [1 + x] ]

Now, we can use the log properties to decompose this into two logs.

y = ln(e^(-x)) + ln(1 + x)

ln and e are inverses of each other, so the first log will leave us with just -x.

y = -x + ln(1 + x)

Now, we differentiate easily. Note that the derivative of ln(x) is equal to 1/x, so

dy/dx = -1 + 1/(1 + x)

If we wanted a cleaner form,

dy/dx = [(-1)(1 + x) + 1] / (1 + x)

And now, simplifying,

dy/dx = [-1 - x + 1] / (1 + x)
dy/dx = -x/(1 + x)

Answer 2

Not sure what the actual problem is here, but assuming you want to find y', this isn't actually logarithmic differentiation per se - you just need to apply the chain rule:

y' = [1/(e^-x + xe^-x)] . [(e^(-x).(-1) + (1.e^-x + x.e^-x.(-1))]
Multiply top and bottom by e^x:
y' = [1 / (1 + x)] . [-1 + 1 + (-x)]
= -x / (x + 1).

You can also write this as -1 + 1 / (x+1) if you prefer.

Answer 3

y = ln|e^-x*(1+x)|= ln|e^-x| + ln|1+x| = -x + ln|1+x|
dy/dx = -1 +1/(1+x)