Question

d/dx (x + 1/x + logx + tan x) =
please give me the correct answer... I need it seriously... ​

Answer 1

Answer:

1 - 1 / x² + 1 / x + sec² x

Step-by-step explanation:

Let :

y = x + 1 / x + log x + tan x

We know :

d / d x ( log x ) = 1 / x

d / d x ( tan x ) = sec² x

d / d x ( 1 / x ) = - 1 / x²

Now diff. w.r.t. :

= > d y / d x = ( x + 1 / x + log x + tan x )'

= > d y / d x = ( 1 - 1 / x² + 1 / x + sec² x )

Therefore , we get required answer!

Answer 2

Answer:

1 - 1 / x² + 1 / x + sec² x

Step-by-step explanation:

Let :

y = x + 1 / x + log x + tan x

We know :

d / d x ( log x ) = 1 / x

d / d x ( tan x ) = sec² x

d / d x ( 1 / x ) = - 1 / x²

Now diff. w.r.t. :

= > d y / d x = ( x + 1 / x + log x + tan x )'

= > d y / d x = ( 1 - 1 / x² + 1 / x + sec² x )

Therefore , we get required answer!

Step-by-step explanation: