Question

differentiate y =
$\sqrt{x } - \frac{1}{ \sqrt{ x} }$
​

Answer 1

Step-by-step explanation:

$\sqrt{x} - \frac{1}{ \sqrt{x} } = y \\ \frac{dy}{dx} = \frac{d}{dx}( {x}^{ \frac{1}{2} } - {x}^{ \frac{ - 1}{2} } ) \\ \: \: = \frac{1}{2} {x}^{ \frac{1}{2} - 1} + - ( - \frac{1}{2} {x}^{ \frac{ - 1}{2} - 1 } ) \\ = \frac{1}{2} {x}^{ \frac{ - 1}{2} } + \frac{1}{2} {x}^{ \frac{ - 3}{2} } \\ = \frac{1}{2 \sqrt{x} } + \frac{1}{ {2}^{ \frac{3}{2} } } \\ = \frac{1}{2 \sqrt{x} } + \frac{1}{2x \sqrt{x} } \\ \frac{dy}{dx} = \frac{1}{2 \sqrt{x} } (1 \: + \frac{1}{x} )$

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