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

Question

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

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Answer

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

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