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

MCQ

If $y=\left(1+\frac{1}{x}\right)^x$, then $\frac{d y}{d x}=$

A
$\left(1+\frac{1}{x}\right)^x \log \left(1+\frac{1}{x}\right)$
B
$\left(x+\frac{1}{x}\right)^x\left\{\log \left(1+\frac{1}{x}\right)+\frac{1}{x+1}\right\}$
✓
$\left(1+\frac{1}{x}\right)^x\left\{\log \left(1+\frac{1}{x}\right)-\frac{1}{x+1}\right\}$
D
$\left(x+\frac{1}{x}\right)^x\left\{\log (x+1)-\frac{x}{x+1}\right\}$

✓

Answer

Correct option: C.

$\left(1+\frac{1}{x}\right)^x\left\{\log \left(1+\frac{1}{x}\right)-\frac{1}{x+1}\right\}$

$y=\left(1+\frac{1}{x}\right)^x$
Taking log on both sides,
$\log y=\log \left(1+\frac{1}{x}\right)^z$
$\log y=x \log \left(1+\frac{1}{x}\right)$
Differentiate with respect to $x,$
$\frac{1}{y} \frac{d y}{d x}=x \times \frac{1}{1+\frac{1}{x}} \times \frac{-1}{x^2}+\log \left(1+\frac{1}{x}\right)$
$\frac{1}{y} \frac{d y}{d x}=\frac{x^2}{x+1} \times \frac{-1}{x^2}+\log \left(1+\frac{1}{x}\right)$
$\frac{d y}{d x}=y\left(\frac{-1}{x+1}+\log \left(1+\frac{1}{x}\right)\right)$
$\frac{d y}{d x}=\left(1+\frac{1}{x}\right)^x\left(\frac{-1}{x-1}+\log \left(1+\frac{1}{x}\right)\right)$
$\frac{d y}{d x}=\left(1+\frac{1}{x}\right)^x\left(\log \left(1+\frac{1}{x}\right)-\frac{1}{x+1}\right)$

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