Question

find d/dx of square root x+1/x-1

Answer 1

Answer:

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

by rationalising the numerator

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

$\frac{d}{dx} (\frac{x+1}{\sqrt{x^{2} -1} } )$

$\frac{d}{dx}(\frac{x}{\sqrt{x^{2} -1} } ) + \frac{d}{dx}(\frac{1}{\sqrt{x^{2} -1} })$

using u/v formula

$\frac{\sqrt{x^{2} -1} -x(\frac{1}{2\sqrt{x^{2} -1}})}{x^{2} -1}$ + $\frac{-1(\frac{1}{2(\sqrt{ {x^{2} -1}) } })}{x^{2} -1}$

$\frac{{\sqrt{x^{2} -1} -(x+1)(\frac{1}{2\sqrt{x^{2} -1}})}{}}{x^{2} -1}$

By taking L.C.M in the numerator

$\frac{2x^{2} -x-3}{x^{2} -1}$