MCQ
જો $y = {(1 + x)^x},$ તો ${{dy} \over {dx}} = $
- A${(1 + x)^x}\left[ {{x \over {1 + x}} + \log ex} \right]$
- B${x \over {1 + x}} + \log (1 + x)$
- ✓${(1 + x)^x}\left[ {{x \over {1 + x}} + \log (1 + x)} \right]$
- Dએકપણ નહીં
Taking log on both sides, $\log y = x\log (1 + x)$
Differentiating w.r.t. $x,$ we get
$\frac{1}{y}\frac{{dy}}{{dx}} = \log (1 + x) + x\frac{1}{{(1 + x)}}$
Thus $\frac{{dy}}{{dx}} = {(1 + x)^x}\left[ {\frac{x}{{1 + x}} + \log (1 + x)} \right]$
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