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Differentiation (p-1) — Maths (commerce) STD 12 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 12 Commerce / ArtsMaths (commerce)Differentiation (p-1)2 Marks
Question
Find $\frac{d y}{d x}$ if, :
$
y=\sqrt{x+\frac{1}{x}}
$
✓

Answer

Given : $y=\sqrt{x+\frac{1}{x}}$
Let $u=x+\frac{1}{x}$
Then $y=\sqrt{u}$
$
\therefore \frac{d y}{d u}=\frac{d}{d u}\left(u^{\frac{1}{2}}\right)=\frac{1}{2} u^{-\frac{1}{2}}
$
$
=\frac{1}{2 \sqrt{u}}=\frac{1}{2 \sqrt{x+\frac{1}{x}}}
$
$
\text { and } \begin{aligned}
\frac{d u}{d x} & =\frac{d}{d x}\left(x+\frac{1}{x}\right) \\
& =\frac{d}{d x}(x)+\frac{d}{d x}\left(x^{-1}\right) \\
& =1+(-1) x^{-2}=1-\frac{1}{x^2}
\end{aligned}
$
$
\begin{aligned}
\therefore \frac{d y}{d x} & =\frac{d y}{d u} \cdot \frac{d u}{d x}=\frac{1}{2 \sqrt{x+\frac{1}{x}}} \cdot\left(1-\frac{1}{x^2}\right) \\
& =\frac{1}{2}\left(x+\frac{1}{x}\right)^{-\frac{1}{2}}\left(1-\frac{1}{x^2}\right) .
\end{aligned}
$

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