Question

derivative of y=ln((e^-x)+x(e^-x))
can you take me through the steps

Answer 1

do i turn it into y=ln((e^-2x)x)

Answer 2

\[\ln(e^{-x}+xe^{-x})\]Is it?

Answer 3

correct

Answer 4

\[\ln(e^{-x}(1+x))\]taking e^{-x} as common.

Answer 5

\[=\ln(e^{-x})+\ln(1+x)\]applying the property ln(ab)=ln(a)+ln(b)

Answer 6

are you there?

Answer 7

yep so far so good

Answer 8

\[=-x+\ln(1+x)......since ......\ln(e^{-x})=\log_{e}(e^{-x})=-x\]

Answer 9

so well have
for the second step
(1/e^-x)(d/dx(e^-x) + (1/1+x)(d/dx(1+x))

Answer 10

Now can you take the derivative here?

Answer 11

Can you take the derivative of -x+ln(1+x) ?

Answer 12

-1+ (1/1+x)d/dx(1+x)

Answer 13

= -1+ 1/1+x

Answer 14

correct, that will be your answer\[\frac{ -x }{ 1+x }\]

Answer 15

that wasnt so bad lol thanks