\[ f(x) = x^x \cdot (\ln x + 1)^{\frac{1}{x}} \]

Question

Find the derivative \( f'(x) \).                              \[ f(x) = x^x \cdot (\ln x + 1)^{\frac{1}{x}} \] 

Answer 1

Given, f(x) = x^x·(lnx + 1)^1/x

Also, \({d(x^x)\over dx} = x^x(lnx + 1) \) and \({d(({lnx} +1)^{1\over x})\over dx} = \left[{1\over (lnx + 1)} - ln(lnx +1)\right]\left((lnx + 1)^{1\over x}\over x^2\right)\)

Then,

 f'(x) = \(d(f(x))\over dx\)= \({d(x^x)\over dx}\cdot (lnx + 1)^{1\over x} + x^x\cdot{d((lnx+1)^{1\over x})\over dx}\)

 = \(x^x(lnx+1)\cdot(lnx+1)^{1\over x} + x^x \cdot\left[{1\over (lnx + 1)} - ln(lnx +1)\right]\left((lnx + 1)^{1\over x}\over x^2\right) \) 

\(\therefore \) f'(x) = \(x^{x-2}\cdot(lnx+1)^{1\over x}\cdot\left[x^2(lnx+1) + {1\over (lnx+1)} - ln(lnx+1)\right] \)