Question

what is (e^x)/(e^-x)? Is it just 1 since the (-x) would just be x since e^-x is the denominator?

Answer 1

no... \[\frac{ e^x }{ e^x } = 1 \cdots \frac{ e^x }{ e^{-x} } = (e^x)(e^x) = (e^x)^2 = ?\]

Answer 2

How would I differentiate ln(e^-x + xe^x)? I know I'm supposed to use both the chain rule and the product rule, somehow...

Answer 3

\[\ln (e^x + xe^x) \text{ Is this your function?}\]

Answer 4

Yeah

Answer 5

\[\frac{ d }{ du }(\ln u)=\frac{ u' }{ u }\]