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

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

#### 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
Is there an error in this question or solution?

#### APPEARS IN

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

Share