Question

find dy\dx,when y=√x-1\√x

Answer 1

Hi...☺

Here is your answer...✌

$y = \frac{ \sqrt{x} - 1}{ \sqrt{x} } \\ \\ y = \frac{ \sqrt{x} }{ \sqrt{x} } - \frac{1}{ \sqrt{x} } \\ \\ y = 1 - \frac{1}{ \sqrt{x} } \\ \\ differentiating \: both \: sides \: w.r.t. \: x \\ we \:get \\ \\ \frac{dy}{dx} = \frac{d}{dx}(1) - \frac{d}{dx} ( \frac{1}{ \sqrt{x} } ) \\ \\ = 0 - \frac{d}{dx} ( {x}^{ \frac{ - 1}{2} } ) \\ \\ = - ( - \frac{1}{2} \: {x}^{(\frac{ - 1}{2} - 1)} ) \\ \\ = \frac{1}{2 } \: {x}^{ (\frac{ - 3}{2}) } \\ \\ = \frac{1}{2 \: {x}^{ \frac{3}{2} } } \: \: \: \: \: \: \: ( \: or \: \frac{1}{2 \sqrt{ {x}^{3} } } \: )$