Question

Differentiate the function with respect to x.

`(x + 1/x)^x + x^((1+1/x))`

Answer 1

Let y = `(x + 1/x)^x + x^((1+1/x))` = u +v

Where u = `(x + 1/x)^x` and v = `x ^((1+1/x))`

Differentiating the above w.r.t. x we get

`dy/dx = (du)/dx + (dv)/dx`  .....(i)

Now, u = `(x + 1/x)^x`

Taking log on both sides, we get,

= `logu = x log (x + 1/x)`   ......(ii)

Differentiating (ii) w.r.t. x, we get

`1/u (du)/dx = x d/dx log (x + 1/x) + log (x + 1/x)(1)`

= `x/(x + 1/x) (1 - 1/x^2) + log (x + 1/x)`

⇒ `(du)/dx = (x + 1/x)^x [x/(x + 1/x)(1 - 1/x^2) + log (x + 1/x)]`  ....(iii)

Also, v = `x^((1 + 1/x))`

Taking log on both sides, we get,

log v = `(1 + 1/x) log x`   ....(iv)

Differentiating (iv) w.r.t. x, we get,

`1/v (dv)/dx = (1 + 1/x)d/dx log x + log x d/dx (1 + 1/x)`

= `(1 + 1/x) 1/x + log x (-1/x^2)`

`(dv)/dx = x^((1+1/x)) [(1 + 1/x) 1/x + log x (-1/x^2)]`  ....(v)

Substituting the value of (iii) and (v) in (i), we get,

`dy/dx = (x + 1/x)^x [x/(x + 1/x) (1 - 1/x^2) + log (x + 1/x)] + x^((1 + 1/x)) [(1 + 1/x) 1/x + log x (-1/x^2)]`