If `y=x^(tanx)+sqrt((x^2+1)/2),fin d(dy)/(dx)`

Options

A. `[(tan x)/(x) + log x . sec^(2) x ] tan x + (x)/(sqrt(2 (x^(2) +1)))`
B. `x^(tan x)[(tan x)/(x) + log x . sec^(2) x ] + (x^(2))/(sqrt(2 (x^(2) +1)))`
C. `x^(tan x)[(tan x)/(x) + log x . sec^(2) x ] + (x)/(sqrt(2 (x^(2) +1)))`
D. None of the above

Answer

C

Explanation

We have , `y = x^(tan x ) + sqrt((x^(2) +1)/(2))`
On taking , `u = x^(tan x ) " and v" + sqrt((x^(2) +1)/(2))`
` log u = tan x log x ` ...(i)
and `v^(2) = (x^(2) +1)/(2) ` ...(ii)
On differentiating Eq. (i) w.r.t.x, we get
`(1)/(u) .(du)/(x) = tanx (1)/(x) + log x . sec^(2) x `
` rArr (du)/(dx) = u [(tanx)/(x) + log x . sec^(2)x]`
` = x^(tanx) [(tanx)/(x) + log x . sec^(2) x]`
Also , differentiating Eq. (ii) w.r.t.x, we get
`2v.(dv)/(dx) = (1)/(2) (2x) rArr (dv)/(dx) = (1)/(4v) .(2x)`
`rArr (dv)/(dx) = (1)/(root(4)((x^(2) +1)/(2))) 2x = (x.sqrt(2))/(2sqrt( x^(2) +1))`
`rArr (dv)/(dx) = (x)/(sqrt(2) (x^(2) +1))`
Now , ` y = u + v `
` therefore (dy)/(dx) = (du)/(dx) + (dv)/(dx)`
`= x^(tanx ) [ (tan x)/(x) + log . sec^(2) x] + (x)/(sqrt(2(x^(2) + 1)))`