If `y=e^(tan x)+(log x)^(tan x)` then find `(dy)/(dx)`

Explanation

Given, `y=e^(tan x) + (log x)^(tan x)`
Let `u=(log x)^(tan x)`
Taking log of both sides,
` log u = (tan x ) * (log log x)`
Differentiating w.r.t.x,
`(1)/(u)*(du)/(dx)=tan x *(1)/(log x)*(1)/(x)+ (log log x) * sec^(2)x`
`:.(du)/(dx)=u[(tan x)/(x*log x)+(log log x)* sec^(2)x]`
`=(log x)^(tan x) [ (tan x)/(x*log x)+(log log x)*sec^(2)x]`
Now, `y=e^(tan x)+u`
`:.(dy)/(dx)=e^(tan x)*sec^(2)x+(du)/(dx)`
`(dy)/(dx) = e^(tan x)*sec^(2)x + (log x)^(tan x)`. `[(tan x)/(x* log x) + (log log x) * sec^(2)x]`