What is the first differential of $Y=(2+tanx)logx$ ?

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Answer 1 · 2017-05-18T05:10:30

$(dY)/dx= (2+tanx)/x+lnxsec^2x$
[Given assumptions stated below]

First a couple of assumptions on this question:
(i) you are looking for the first derivative $((dY)/dx)$
(ii) $log$ is natural log ( $ln$ )

Then, applying the Product Rule and standard differentials for $lnx$ and $tanx$ :

$(dY)/dx= (2+tanx)* d/dx(lnx) + lnx* d/dx(2+tanx)$

$= (2+tanx)* 1/x+lnx*(0+sec^2x)$

$= (2+tanx)/x+lnxsec^2x$

What is the first differential of Y=(2+tanx)logxY=(2+tanx)logx ?