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Question
Mark the correct alternative in of the following:
If \[y = \sqrt{x} + \frac{1}{\sqrt{x}}\] then \[\frac{dy}{dx}\] at x = 1 is
Options
1
\[\frac{1}{2}\]
\[\frac{1}{\sqrt{2}}\]
0
Solution
\[y = \sqrt{x} + \frac{1}{\sqrt{x}}\]
\[ = x^\frac{1}{2} + x^{- \frac{1}{2}}\]
Differentiating both sides with respect to x, we get
\[\frac{dy}{dx} = \frac{d}{dx}\left( x^\frac{1}{2} + x^{- \frac{1}{2}} \right)\]
\[ = \frac{d}{dx}\left( x^\frac{1}{2} \right) + \frac{d}{dx}\left( x^{- \frac{1}{2}} \right)\]
\[ = \frac{1}{2} x^\frac{1}{2} - 1 + \left( - \frac{1}{2} \right) x^{- \frac{1}{2} - 1} \left( y = x^n \Rightarrow \frac{dy}{dx} = n x^{n - 1} \right)\]
\[ = \frac{1}{2} x^{- \frac{1}{2}} - \frac{1}{2} x^{- \frac{3}{2}}\]
Putting x = 1, we get
\[\left( \frac{dy}{dx} \right)_{x = 1} = \frac{1}{2} \times 1 - \frac{1}{2} \times 1 = 0\]
Thus, \[\frac{dy}{dx}\] 1 is 0.
Hence, the correct answer is option (d).
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