Question

Differentiate the function.
y = ln(e^−x + xe^−x)

Answer 1

\[\frac{d}{dx}[\ln(g(x))]=\frac{g'(x)}{g(x)}\]

Answer 2

That doesn't help me a lot haha. i don't know the derivative of (e^−x + xe^−x)

Answer 3

ooh ok product and chain rule for this

Answer 4

\[\frac{d}{dx}e^{-x}=-e^{-x}\] by the chain rule

Answer 5

then
\[\frac{d}{dx}xe^{-x}=e^{-x}-xe^{-x}\] by the product rule

Answer 6

put them together in the numerator, you are left only with
\[-xe^{-x}\]

Answer 7

Thank you so much!

Answer 8

yw