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Mark the Correct Alternative in of the Following: If Y = √ X + 1 √ X Then D Y D X at X = 1 is - Mathematics

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MCQ

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

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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).

 

 

Concept: The Concept of Derivative - Algebra of Derivative of Functions
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 30 Derivatives
Q 6 | Page 48

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