Question

What is the derivative if this ..... .
lnx +e^x / tanx

Answer 1

Answer:

( tan² x + e^x ( tan x - sec² x ) ) / x . tan² x

Step-by-step explanation:

Let :

y = ㏑ x + e^x / tan x

We know :

( e^x )' = e^x

( ㏑ x )' = 1 / x

( tan x )' = sec² x

Diff. w.r.t. x :

= > d y / d x = 1 / x + ( tan x ( e^x )' - e^x ( tan x )' ] / tan² x

= > d y / d x = 1 / x + e^x . tan x - e^x . sec² x ) / tan² x

= > d y / d x = ( tan² x + e^x ( tan x - sec² x ) ) / x . tan² x

Hence we get required answer!

Answer 2

Step-by-step explanation:

Answer✍️

• ( tan² x + e^x ( tan x - sec² x ) ) / x . tan² x

Let :

• y = ㏑ x + e^x / tan x

We know :

• ( e^x )' = e^x

• ( ㏑ x )' = 1 / x

• ( tan x )' = sec² x

Diff. w.r.t. x :

➡️d y / d x = 1 / x + ( tan x ( e^x )' - e^x ( tan x )' ] / tan² x

➡️d y / d x = 1 / x + e^x . tan x - e^x . sec² x ) / tan² x

➡️d y / d x = ( tan² x + e^x ( tan x - sec² x ) ) / x . tan² x

hope this helps you