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

Answer

N/a

Explanation

We have `y=x^(tanx)+sqrt((x^(2)+1)/(2))"...."(i)`.
Taking ` u=x^(tanx)` and `v = sqrt((x^(2)+1)/(2))`
`logu = tanxlogx"...."(ii)`
and `v^(2) = (x^(2)+1)/(2)"....."(iii)`
On differentiating Eq. (ii) w.r.t.x, we get
`1/u.(du)/(dx)= tanx.(1)/(x)+logx.sec^(2)x`
`rArr (du)/(dx)=u[(tanx)/(x)+logx.sec^(2)x]`
`=x^(tanx)[(tanx)/(x)+logx.sec^(2)x]"....."(iv)`
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)/(4.sqrt((x^(2)+1)/(2))).2x= (x.sqrt(2))/(2sqrt(x^(2)+1))`
`rArr (dv)/(dx)=(x)/(sqrt(2(x^(2)+1)))"...."(v)`
Now, `y=u+v`
`:. (dy)/(dx)=(du)/(dx)+(dv)/(dx)`
`=x^(tanx)[(tanx)/(x)+logx.sec^(2)x]+(x)/(sqrt(2(x^(2)+1)))`