Question

Question #1b978

Answer 1

`1-tanx`

Explanation:

Assumung `log` is in base `e`, `logbeta equiv lnbeta` for the basis of this quesion:

`log(e^x cosx )` = `log e^x + log(cosx)`

`=x+log(cosx)`
`d/dx ( x) = 1`
`d/dx ( 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`