Question

differentiate the following function​

Answer 1

Answer:

Let's assume the given expression be y.

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

Can be rewritten as,

${y =\left( \dfrac{\sqrt{x+1} + \sqrt{x-1}}{\sqrt{x+1} - \sqrt{x-1}}\right)\cdot \left(\dfrac{\sqrt{x+1}+ \sqrt{x-1}}{\sqrt{x+1} +\sqrt{x-1}}\right)}$

${y =\left( \dfrac{(\sqrt{x+1} + \sqrt{x-1})(\sqrt{x+1} +\sqrt{x-1})}{(\sqrt{x+1} - \sqrt{x-1})(\sqrt{x+1} +\sqrt{x-1})}\right)}$

${y = \dfrac{(\sqrt{x+1} + \sqrt{x-1})^2}{(\sqrt{x+1})^2 - (\sqrt{x-1})^2}}$

${y = \dfrac{(\sqrt{x+1})^2 + (\sqrt{x-1})^2 + 2\sqrt{(x+1)(x-1)}}{x + 1 -x + 1 }}$

${y = \dfrac{x+1 + x-1 + 2\sqrt{(x+1)(x-1)}}{2}}$

${y = \dfrac{2x + 2\sqrt{(x+1)(x-1)}}{2}}$

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

Now differentiating both sides w.r.t. x

${\dfrac{dy}{dx} = \dfrac{d}{dx}\Big(x +\sqrt{(x+1)(x-1)}\Big)}$

${\dfrac{dy}{dx} = \dfrac{d}{dx}\Big(x\Big) +\dfrac{d}{dx}\Big(\sqrt{(x+1)(x-1)}\Big)}$

Now, we will use the following rules,

${\bullet\qquad \dfrac{d}{dx} x^n = n\ x^{n-1}}$

${\bullet\qquad \dfrac{d}{dx} \sqrt{x} = \dfrac{1}{2\sqrt{x}}}$

${\bullet\qquad \dfrac{d}{dx} f(g(x)) = f'(g(x))\ g'(x)}$

${\bullet\qquad \dfrac{d}{dx} f(x) \cdot g(x) = f'(x) g(x) + f(x) g'(x)}$

Using these formulas, we get:
${\dfrac{dy}{dx} = 1 +\dfrac{1}{2\sqrt{(x+1)(x-1)}}\cdot \dfrac{d}{dx}\Big((x+1)(x-1)\Big)}$

${\dfrac{dy}{dx} = 1 +\dfrac{1}{2\sqrt{x^2-1}}\cdot\Big((x+1) + (x-1)\Big)}$

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

$\boxed{\dfrac{dy}{dx} = 1 +\dfrac{x} {\sqrt{x^2-1} } }$

This is the required answer.