Correct Answer - `e^(x sin x^(3))(3x^(3) cos x^(3) + sin x^(3))+(tan x) ^(x) [2x " cosec " 2x + log (tan x)]`
Since, ` y = e^(x sin x^(3))+(tan x)^(x),` then
` y = u + v," where " u=e^(x sin x ^(3)) and v = (tan x ) ^(x) `
` rArr" " (dy)/(dx) = ((du)/(dx)+(dv)/(dx))` …(i)
Here, ` u = e^(x sin x^(3)) ` and log v = x log (tan x)
On differentiating both sides w.r.t. x, we get
`(du)/(dx) = e^(x sin x^(3))* (3x^(3) cos x^(3) + sin x^(3))` ..(ii)
and ` 1/v*(dv)/(dx) = (x*sec^(2) x)/(tan x) + log (tan x) `
`(dv)/(dx) = (tan x)^(x) [2x* cosec (2x)+ log (tan x)}`...(iii)
From Eqs. (i), (ii)and (iii), we get
` (dy)/(dx) = e^(x sin x^(3))(3x^(3)*cos x^(3) + sin x^(3)) + (tan x)^(x) `
[2x cosec 2x + log (tan x)]
