Question

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

Answer 1

`(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`