Question

Derive this please

Answer 1

\[\log_{2}(e ^{-x} \cos Pix)\]

Answer 2

what's your question?

Answer 3

find the derivative of that expression

Answer 4

"differentiate"

Answer 5

yes, "Differentiate" the terms are interchangeable.

Answer 6

can you help or not?

Answer 7

pretty demanding.

Answer 8

You're just trying to "correct" me when I'm not wrong, instead of helping with the problem I put up. That's all

Answer 9

because you 'put up' something that was nonsense, so no one is going to help you.
except me, because I'm nice enough to see if there's a real problem you're working with...
convert the base to e
rewrite ln(a*b) as ln(a) + ln(b)
\[\frac{ \ln e^{-x} +x ln(\cos \pi x) }{ \ln2 }\]

\[\frac{ -x+ ln(\cos \pi x) }{ \ln2 }\]
\[(\ln u) ' = \frac{ du }{ u} = \frac{-\pi \sin \pi x}{ \cos \pi x } = -\pi \tan x\]
so:
\[\frac{-1 -\pi*\tan(\pi x) }{ \ln2 }\]

Answer 10

\[\frac{ \ln e^{-x} +\ln(\cos \pi x ) }{ \ln 2 }\]
\[\frac{ -x +\ln(\cos \pi x ) }{ \ln 2 }\]

\[(\frac{ \ln \cos \pi x }{ \ln 2 }) ' => \frac{ du }{ u } = \frac{ - \pi*\sin \pi x }{\ln 2 * \cos \pi x }\]
\[= \frac{ - \pi*\tan \pi x }{\ln 2 }\]
so:
\[\frac{ -1-\pi \tan \pi x }{ \ln 2 }\]

Answer 11

edited for two typos

Answer 12

"Derive" and "differentiate" are not interchangeable.