Question #1b978

1 Answer
Dec 4, 2017

1-tanx 1tanx

Explanation:

Assumung loglog is in base ee, logbeta equiv lnbeta logβlnβ for the basis of this quesion:

log(e^x cosx )log(excosx) = log e^x + log(cosx)logex+log(cosx)

=x+log(cosx)=x+log(cosx)
d/dx ( x) = 1ddx(x)=1
d/dx ( log(cosx) ) ddx(log(cosx));

We know d/dx ( log(gamma(x))) = (gamma'(x))/(gamma(x)

So d/dx ( log(cosx)) = 1-sinx / cosx = 1-tanx

This answer differs for log_10 but is simple for base e