Question

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

Find `(dy)/(dx), if y = x^tanx + sqrt(x^2 + 1)/2`.

Answer 1

`y = x^tanx + sqrt(x^2 + 1)/2    ...(i)`

Let u = `x^tanx, v = sqrt(x^2 + 1)/2`

Equation (i) becomes

y = u + v

Differentiate w.r.t. x,

`(dy)/(dx) = (du)/(dx) + (dv)/(dx)`

Now, u = x^{tan x}

taking log on both sides

log u = log (x^{tan x})

log u = tan x log x

Differentiate w.r.t. x,

`1/u (du)/(dx) = (tan x)/x + log x  sec2^2 x`

⇒ `(du)/(dx) = u[(tan x)/x  + log x  sec^2 x]`

⇒ `(du)/(dx) = x^tan x[(tan x)/x + log x  sec^2 x]`

Again take,

v = `sqrt(x^2 - 1)/2`

`(dv)/(dx) = 1/2 xx 1/(2sqrt(x^2 + 1)) xx (2x)`

= `x/(2sqrt(x^2 + 1)`

Put the value `(du)/(dx) and (dv)/(dx)  "in equation" (ii),`

Now, `(dy)/(dx) = x^tanx[tan x/x + log x  sec^2 x] + x/(2sqrt(x^2 + 1)`