Question

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

Answer 1

HI

Answer 2

hi

Answer 3

ive made a lot of progress on this and get stuck a bit through it

Answer 4

\[\ln (e^-x(1+x)) = \ln(e^-x) + \ln(1+x) = \frac{ 1 }{ e^-x } + \frac{ 1 }{ (x+1) }\]

Answer 5

how is ln(e^-x) = -x?

Answer 6

a little typo when u derivate\[\ln(e^{-x})\]

Answer 7

ahh yeah thats it

Answer 8

ya im not sure why its doing that

Answer 9

how is it -x?

Answer 10

e^x is inverse of ln x

Answer 11

hmmm

Answer 12

so ln(e^-x) which is 1/e^-x yields -x somehow

Answer 13

\[\frac{d}{dx}\ln(f(x))=\frac{f'(x)}{f(x)}\] is what you need

Answer 14

\[f(x)=e^{-x}+xe^{-x}\]
\[f'(x)=-e^{-x}+e^{-x}-xe^{-x}\] by the chain and product rule
stick that in the numerator

Answer 15

well you might want to write
\[f'(x)=-xe^{-x}\] since the first two terms add up to zero

Answer 16

lets start at the beginning then

Answer 17

ok, ln(e^-x + xe^-x)

Answer 18

my understanding is, you want the derivative of
\[\ln(e^{-x}+xe^{-x})\] is that correct?

Answer 19

i could factor out that e^-x

Answer 20

ok
then the derivative of the log of something, is the derivative of the something, over the something
more succinctly written as
\[\frac{d}{dx}\ln(f(x))=\frac{f'(x)}{f(x)}\]

Answer 21

ln(x) = 1/x

Answer 22

i wouldn't factor, that will just make more work, although i guess you could
i would just go ahead and take the derivative of \(f(x)=e^{-x}+xe^{-x}\) and then put the derivative over \(f(x)\)

Answer 23

by the way it is certainly not true that \(\ln(x)=\frac{1}{x}\) what you mean is that
\[\frac{d}{dx}\ln(x)=\frac{1}{x}\] so by the chain rule
\[\frac{d}{dx}\ln(f(x))=\frac{f'(x)}{f(x)}\]

Answer 24

in other words, this is your job:
take the derivative of the "inside function" namely \(f(x)=e^{-x}+xe^{-x}\) and then divide that result by \(f(x)=e^{-x}+xe^{-x}\) itself
you are looking for \(\frac{f'(x)}{f(x)}\) as an answer

Answer 25

for example
\[\frac{d}{dx}\ln(\sin(x))=\frac{\cos(x)}{\sin(x)}\]

Answer 26

I guess my question is how is ln(e^-x) = -x

Answer 27

\[\frac{d}{dx}\ln(x^2+3x)=\frac{2x+3}{x^2+3x}\] etc

Answer 28

because logs and exponentials are inverse fuctions, so it is always the case that
\[\log_b(b^x)=x\] and \[b^{\log_b(x)}=x\]

Answer 29

ok, perfect, thank you!

Answer 30

yw