Find d y d x if, : y= (1+1 x )^x - Vidyadip

Question

Find $\frac{d y}{d x}$ if, :
$
y=\left(1+\frac{1}{x}\right)^x
$

✓

Answer

$
\begin{aligned}
& y=\left(1+\frac{1}{x}\right)^x \\
& \therefore \log y=\log \left(1+\frac{1}{x}\right)^x=x \log \left(1+\frac{1}{x}\right)
\end{aligned}
$
Differentiating both sides w.r.t. $x$, we get
$
\begin{aligned}
\frac{1}{y} \cdot \frac{d y}{d x} & =\frac{d}{d x}\left[x \log \left(1+\frac{1}{x}\right)\right] \\
& =x \frac{d}{d x}\left[\log \left(1+\frac{1}{x}\right)\right]+\left[\log \left(1+\frac{1}{x}\right)\right] \cdot \frac{d}{d x}(x) \\
& =x \times \frac{1}{1+\frac{1}{x} \cdot \frac{d}{d x}\left(1+\frac{1}{x}\right)+\left[\log \left(1+\frac{1}{x}\right)\right] \times 1} \\
& =x \times \frac{x}{x+1} \times\left(0-\frac{1}{x^2}\right)+\log \left(1+\frac{1}{x}\right) \\
\therefore \frac{d y}{d x} & =y\left[\frac{-1}{x+1}+\log \left(1+\frac{1}{x}\right)\right] \\
& =\left(1+\frac{1}{x}\right)^x\left[\log \left(1+\frac{1}{x}\right)-\frac{1}{1+x}\right]
\end{aligned}
$

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