MCQ
If $y = \log {x^x},$ then ${{dy} \over {dx}} = $
- A${x^x}(1 + \log x)$
- ✓$\log (ex)$
- C$\log \left( {{e \over x}} \right)$
- DNone of these
Differentiating w.r.t. $x,$ we get
$\frac{{dy}}{{dx}} = (1 + \log x) = \log e + \log x = \log (ex)$, $(\because \log e = 1)$
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$(A)$ $ \equiv \frac{{x + 1}}{1} = \frac{{y - 2}}{{ - 2}} = \frac{{z - 0}}{1}$
$(B)$ $ \equiv \frac{x}{1} = \frac{y}{{ - 2}} = \frac{{z - 1}}{1}$
$(C)$ $ \frac{{x + 1/2}}{1} = \frac{{y - 1}}{{ - 2}} = \frac{{z - 1/2}}{1}$
$(i)$ At which point, $Z$ is minimum?
$(ii)$ At which point, $Z$ is maximum ?
$(iii)$ The maximum value of $\mathrm{Z}$ is $\ldots \ldots \ldots$
$(iv)$ The minimum value of $\mathrm{Z}$ is $\ldots \ldots \ldots$