Find d y d x if : y= (x+1 x )^2

Question

Find $\frac{d y}{d x}$ if : $\mathrm{y}=\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2$

✓

Answer

$
\begin{aligned}
y & =\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2 \\
\therefore \quad y & =x+2+\frac{1}{x} \text { }
\end{aligned}
$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d}}{\mathrm{d} x}\left(x+2+\frac{1}{x}\right) \\
= & \frac{\mathrm{d}}{\mathrm{d} x}(x)+\frac{\mathrm{d}}{\mathrm{d} x}(2)+\frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{1}{x}\right) \\
= & 1+0+\frac{\mathrm{d}}{\mathrm{d} x}\left(x^{-1}\right) \\
= & 1+(-1) x^{-2} \\
= & 1-\frac{1}{x^2} \text { }
\end{aligned}
$

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