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

1 Answer
Alan N.
May 18, 2017

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

Explanation:

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#

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