Question

Find dy/dx when y = √ x - 1/√ x
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Answer 1

$\underline{\huge\sf\purple{Question}}$

Find dy/dx when y = √ x - 1/√ x

$\underline{\huge\sf\purple{Answer}}$

$\bold{= \frac{1}{2\sqrt{x} }( \frac{1+x}{x}) }$

$\underline{\huge\sf\red{Solution}}$

$\bold{\frac{dy}{dx} } = \sf{ \frac{d}{dx} ( \sqrt{x} - \frac{1}{\sqrt{x} } )}$

$\sf{ = \frac{d}{dx} (\sqrt{x} ) - \frac{d}{dx} ( \frac{1}{\sqrt{x} }) }$

$\sf{ = \frac{d}{dx} (x)^{\frac{1}{2}} - \frac{d}{dx} (x)^{-\frac{1}{2}} }$

$\sf{ = \frac{1}{2} \times x^{\frac{1}{2} -1} - [ -\frac{1}{2} \times x^{-\frac{1}{2} - 1 } ] }$

$\sf{=\frac{x^{\frac{1}{2} -1}}{2} + \frac{x^{-\frac{1}{2} - 1}}{2} }$

$\sf{ = \frac{x^{\frac{1 - 2 }{2}}}{2} + \frac{x^{\frac{-1-2}{2}}}{2} }$

$\sf{ = \frac{x^{\frac{-1}{2}}}{2} + \frac{x^{\frac{-3}{2}}}{2} }$

$\sf{ = \frac{\frac{1}{\sqrt{x} }}{2} + \frac{\frac{1}{_{3}\sqrt{x} }}{2} }$

$\sf{ = \frac{1}{2\sqrt{x} } + \frac{1}{2 \ _{3}\sqrt{x} } }$

$\sf{ = \frac{1}{2\sqrt{x} } ( 1 + \frac{1}{x} ) }$

$\sf{= \frac{1}{2\sqrt{x} }( \frac{1+x}{x}) }$

$\huge\sf\red{\colorbox{orange}{= \frac{1}{2\sqrt{x} }( \frac{1+x}{x}) }}$