Question

Ques - y = log ( √x-1 - √x+1 ) . Find dy/dx .

Answer 1

Step-by-step explanation:

Given ,

$y = log( \sqrt{x - 1} - \sqrt{x + 1} )$

Differentiating both sides,

$\frac{dy}{dx} = \frac{ \frac{dy}{dx} ( \sqrt{x - 1} - \sqrt{x + 1} ) }{ \sqrt{x - 1} - \sqrt{x + 1} }$

$= > \frac{dy}{dx} = \frac{ \frac{1}{2 \sqrt{x - 1} } - \frac{1}{2 \sqrt{x + 1} } }{ \sqrt{x - 1} - \sqrt{x + 1} }$

$= > \frac{dy}{dx} = \frac{1}{2}. \frac{ \frac{ \sqrt{x + 1} - \sqrt{x - 1} }{ \sqrt{ {x}^{2} - 1 } } }{ \sqrt{x - 1} - \sqrt{x + 1} }$

$= > \frac{dy}{dx} = \frac{ - 1}{2 \sqrt{ {x}^{2} - 1} }$

Hope this will help you