# Question #8b077

Oct 22, 2014

$y ' = \frac{1}{2 \cdot \sqrt{x}} \cdot \left(\ln \left(x {e}^{x}\right) + 2 + 2 x\right)$

Explanation :

$y = {x}^{\frac{1}{2}} \cdot \ln \left(x {e}^{x}\right)$

Using Product Rule, which is

$y = f \left(x\right) \cdot g \left(x\right)$

differentiating with respect to $x$,

$y ' = f ' \left(x\right) \cdot g \left(x\right) + f \left(x\right) \cdot g ' \left(x\right)$

Now similarly following for the given problem,

$y ' = \left({x}^{\frac{1}{2}}\right) ' \ln \left(x {e}^{x}\right) + {x}^{\frac{1}{2}} \left(\ln \left(x {e}^{x}\right)\right) '$

$y ' = \frac{1}{2} {x}^{- \frac{1}{2}} \cdot \ln \left(x {e}^{x}\right) + {x}^{\frac{1}{2}} \cdot \frac{1}{x {e}^{x}} \left(x {e}^{x}\right) '$

$y ' = \frac{1}{2} \cdot \frac{1}{x} ^ \left(\frac{1}{2}\right) \cdot \ln \left(x {e}^{x}\right) + {x}^{\frac{1}{2}} \cdot \frac{1}{x {e}^{x}} \left({e}^{x} + x {e}^{x}\right)$

$y ' = \frac{1}{2} \cdot \frac{1}{x} ^ \left(\frac{1}{2}\right) \cdot \ln \left(x {e}^{x}\right) + \frac{1}{{x}^{\frac{1}{2}} {e}^{x}} \left({e}^{x} + x {e}^{x}\right)$

$y ' = \frac{1}{2} \cdot \frac{1}{x} ^ \left(\frac{1}{2}\right) \cdot \ln \left(x {e}^{x}\right) + \frac{1}{{x}^{\frac{1}{2}}} \left(1 + x\right)$

$y ' = \frac{1}{2} \cdot \frac{1}{x} ^ \left(\frac{1}{2}\right) \cdot \left(\ln \left(x {e}^{x}\right) + 2 + 2 x\right)$

$y ' = \frac{1}{2 \cdot \sqrt{x}} \cdot \left(\ln \left(x {e}^{x}\right) + 2 + 2 x\right)$