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Question
Differentiate the function with respect to x.
`(x + 1/x)^x + x^((1+1/x))`
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Solution
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)]`
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