If y = (1 + x)^x , then dy dx = - Vidyadip

MCQ

If $y = {(1 + x)^x},$ then ${{dy} \over {dx}} = $

A
${(1 + x)^x}\left[ {{x \over {1 + x}} + \log ex} \right]$
B
${x \over {1 + x}} + \log (1 + x)$
✓
${(1 + x)^x}\left[ {{x \over {1 + x}} + \log (1 + x)} \right]$
D
None of these

✓

Answer

Correct option: C.

${(1 + x)^x}\left[ {{x \over {1 + x}} + \log (1 + x)} \right]$

c
(c) $y = {(1 + x)^x}$

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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