Question

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

Answer 1

oops i had to edit question that might help lol

Answer 2

\[\Large\bf\sf y=\ln\left[e^{-x}-xe^{-x}\right]\]Factoring e^-x out of each term,\[\Large\bf\sf y=\ln\left[e^{-x}(1-x)\right]\]From here we want to apply a rule of logs.\[\Large\bf\sf \log(a\cdot b)\quad=\quad \log(a)+\log(b)\]

Answer 3

\[\Large\bf\sf y=\ln e^{-x}+\ln(1-x)\]Gives us something like that, yes?

Answer 4

Then we'll want to apply another rule of logs before we take a derivative.

Answer 5

\[\Large\bf\sf \log(a^b)\quad=\quad b\cdot \log(a)\]

Answer 6

for the -x?

Answer 7

Yes.

Answer 8

so i got y= -xln(e) + ln(1-x)
y'= -x/e + 1/1-x

Answer 9

Woops, no no don't apply any special derivative rules to ln(e).
It is just a `constant`.

See how there is no variable inside of the log?
It's just a fancy fancy constant wearing a suit, trying to trick you.

Answer 10

Furthermore, you should make this simplification before differentiating.\[\Large\bf\sf \ln(e)\quad=\quad 1\]

Answer 11

oh i see ok so it would just be -x + 1/1-x

Answer 12

i was wondering why there wasnt an e in the answer

Answer 13

You should get -x BEFORE taking a derivative, yes?

Answer 14

That first term should change a lil bit.

Answer 15

yes then it would become a -1?

Answer 16

Yes, good.
One more thing to be careful about.
Need to apply the `chain rule` to the second term.

Answer 17

wouldn't that just be 1/1-x ?

Answer 18

oh nvm 1/1-x * -1

Answer 19

Ya there ya go \c:/

Answer 20

thanks