Question

Differentiate wrt x...

Answer 1

Step-by-step explanation:

HOPE IT'S HELPFUL

THANK YOU...

Answer 2

$</p><p>\sf \rightarrow \quad y = \dfrac{ \sqrt{x} + 1 }{ \sqrt{x} - 1 } \\ \\ \sf \rightarrow \quad y = \dfrac{ \sqrt{x} + 1}{ \sqrt{x} - 1 } \times \dfrac{ \sqrt{x} + 1}{ \sqrt{x} + 1 } \\ \\ \sf \rightarrow \quad y = \dfrac{x + 1 + 2 \sqrt{x} }{x - 1} \\ \\ \sf \rightarrow \quad \dfrac{dy}{dx} = \dfrac{d}{dx} \bigg(\dfrac{x + 1 + 2 \sqrt{x} }{x - 1} \bigg) \\ \\ \sf \rightarrow \quad \dfrac{dy}{dx} = \dfrac{ \dfrac{d}{dx} \bigg(x + 1 + 2 \sqrt{x} \bigg) \cdot (x - 1) - \dfrac{d}{dx} \bigg( x - 1\bigg) \cdot (x + 1 + 2 \sqrt{x} )}{ {(x - 1)}^{2} } \\ \\ \sf \rightarrow \quad \frac{dy}{dx} = \dfrac{ \bigg(1 + \dfrac{ \cancel{2}}{ \cancel{2} \sqrt{x} } \bigg)(x - 1) - (x + 1 + 2 \sqrt{x}) }{ {(x - 1)}^{2} } \\ \\ \sf \rightarrow \quad \frac{dy}{dx} = \dfrac{ \bigg( \dfrac{x + \sqrt{x} }{x} \bigg)(x - 1) -( x + 1 + 2 \sqrt{x}) }{ {(x - 1)}^{2} } \\ \\ \sf \rightarrow \quad \frac{dy}{dx} = \dfrac{ \bigg(\dfrac{ {x}^{2} - x + x \sqrt{x} - \sqrt{x} }{x} \bigg) - (x + 1 + 2 \sqrt{x} )}{ {(x - 1)}^{2} } \\ \\ \sf \rightarrow \quad \frac{dy}{dx} = \dfrac{ \dfrac{ \cancel{ {x}^{2} } - x + x \sqrt{x} - \sqrt{x} - \cancel{ {x}^{2}} - x - 2x \sqrt{x} }{x} }{ {(x - 1)}^{2} } \\ \\ \sf \rightarrow \quad \frac{dy}{dx} = \dfrac{ - 2x + (x - 1 - 2x) \sqrt{x} }{x {(x - 1)}^{2} } \\ \\ \sf \rightarrow \quad \frac{dy}{dx} = \dfrac{ - 2x - (x + 1) \sqrt{x} }{x {(x - 1)}^{2} } </p><p>$