MCQ
જો $y = \log {x^x},$ તો ${{dy} \over {dx}} = $
- A${x^x}(1 + \log x)$
- ✓$\log (ex)$
- C$\log \left( {{e \over x}} \right)$
- Dએકપણ નહીં
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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$f(x)=\left\{\begin{array}{ll}\frac{x^{3}}{(1-\cos 2 x)^{2}} \log _{e}\left(\frac{1+2 x e^{-2 x}}{\left(1-x e^{-x}\right)^{2}}\right), & x \neq 0 \\ \,\alpha & , x=0\end{array}\right.$ જો $\mathrm{f}$ એ $\mathrm{x}=0$ આગળ સતત હોય તો $\alpha$ ની કિમંત મેળવો.