Question

differentiate the following functions with respect to X
$y = \frac{ \sqrt{ x}+1 }{ \sqrt{x - 1} }$
Its a JEE qn. please answer!!!​

Answer 1

Step-by-step explanation:

We have,

$y = \frac{ \sqrt{ x}+1 }{ \sqrt{x - 1} }$

$\implies \: \frac{dy}{dx} = \frac{d}{dx} \bigg(\frac{ \sqrt{ x}+1 }{ \sqrt{x - 1} } \bigg)$

$\implies \: \frac{dy}{dx} =\frac{ \sqrt{x - 1} . \dfrac{d}{dx} (\sqrt{ x}+1) - ( \sqrt{x} + 1). \dfrac{d}{dx} ( \sqrt{x - 1} ) }{ (\sqrt{x - 1} ) ^{2} }$

$\implies \: \frac{dy}{dx} =\frac{ \sqrt{x - 1} . \dfrac{1}{2 \sqrt{x} } - ( \sqrt{x} + 1). \dfrac{1}{2 \sqrt{x - 1} } }{ (\sqrt{x - 1} ) ^{2} }$

$\implies \: \frac{dy}{dx} =\frac{ \dfrac{\sqrt{x - 1}}{2 \sqrt{x} } - \dfrac{ \sqrt{x} + 1 }{2 \sqrt{x - 1} } }{ (\sqrt{x - 1} ) ^{2} }$

$\implies \: \frac{dy}{dx} =\frac{ \dfrac{ ( \sqrt{x - 1})^{2} - \sqrt{x} ( \sqrt{x} + 1 )}{2. \sqrt{x}. \sqrt{x - 1} } }{ (\sqrt{x - 1} ) ^{2} }$

$\implies \: \frac{dy}{dx} =\frac{ \dfrac{ x - 1- ( x + \sqrt{x} )}{2. \sqrt{x}. \sqrt{x - 1} } }{ (\sqrt{x - 1} ) ^{2} }$

$\implies \: \frac{dy}{dx} =\frac{ \dfrac{ x - 1- x - \sqrt{x} }{2. \sqrt{x}. \sqrt{x - 1} } }{ (\sqrt{x - 1} ) ^{2} }$

$\implies \: \frac{dy}{dx} =\frac{ \dfrac{ - 1 - \sqrt{x} }{2. \sqrt{x}. \sqrt{x - 1} } }{ (\sqrt{x - 1} ) ^{2} }$

$\implies \: \frac{dy}{dx} =\frac{ \dfrac{ - ( \sqrt{x} + 1)}{2 \sqrt{x^{2} - x} } }{ (\sqrt{x - 1} ) ^{2} }$

$\implies \: \frac{dy}{dx} =\frac{ \dfrac{ - ( \sqrt{x} + 1)}{2 \sqrt{x^{2} - x} } }{x - 1 }$

$\implies \: \frac{dy}{dx} =\dfrac{ - ( \sqrt{x} + 1)}{2 \sqrt{x^{2} - x} (x - 1 ) }$