Question

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

Answer 1

Given that: y = `x^tanx + sqrt((x^2 + 1)/2)`

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

∴ y = u + v

Differentiating both sides w.r.t. x

`"dy"/"dx" = "du"/"dx" + "dv"/"dx"`   .....(i)

Now taking u = `x^tanx`

Taking log on both sides log u = `log(x^tanx)`

log u = tan x . log x

Differentiating both sides w.r.t. x

`1/"u" * "du"/"dx" = "d"/"dx"(tan x * log x)`

⇒ `1/"u" * "du"/"dx" = tan x * "d"/"dx" (log x) + log x * "d"/"dx" (tan x)`

⇒ `1/"u" * "du"/"dx" = tan x * 1/x + log x * sec^2x`

⇒ `"du"/"dx" = "u"[tanx/x + log x * sec^2x]`

∴ `"du"/"dx" = x^tanx [tanx/x + log x sec^2x]`

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

⇒ v = `1/sqrt(2) sqrt(x^2 + 1)`

Differentiating both sides w.r.t. x

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

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

Putting the values of `"du"/"dx"` and `"dv"/"dx"` in equation (i)

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