Question

$\sqrt{ {x }^{2} + \sqrt{ {x}^{2} + 1} }$
derivatives please make it fast​

Answer 1

Answer:

$\displaystyle{\boxed{\red{\sf\:\dfrac{dy}{dx}\:=\:\dfrac{2x\:+\:\dfrac{x}{\sqrt{x^2\:+\:1}}}{2\:\sqrt{x^2\:+\:\sqrt{x^2\:+\:1}}}}}}$

Step-by-step-explanation:

We have given a function.

We have to find the derivative of that function.

The given function is

$\displaystyle{\sf\:\sqrt{x^2\:+\:\sqrt{x^2\:+\:1}}}$

Let this function be y.

$\displaystyle{\therefore\:\sf\:y\:=\:\sqrt{x^2\:+\:\sqrt{x^2\:+\:1}}}$

$\displaystyle{\implies\sf\:y\:=\:\left(\:x^2\:+\:\sqrt{x^2\:+\:1}\:\right)^{\frac{1}{2}}}$

Differentiating both sides w.r.t x, we get,

$\displaystyle{\sf\:\dfrac{d}{dx}\:(\:y\:)\:=\:\dfrac{d}{dx}\:\left[\:\left(\:x^2\:+\:\sqrt{x^2\:+\:1}\:\right)^{\frac{1}{2}}\:\right]}$

We know chain rule of differentiation,

$\displaystyle{\boxed{\blue{\sf\:\dfrac{d}{dx}\:[\:f\:(\:g\:(\:x\:)\:)\:]\:=\:\dfrac{d}{dx}\:[\:f\:(\:g\:(\:x\:)\:)\:]\:.\:\dfrac{d}{dx}\:[\:g\:(\:x\:)\:]\:}}}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:\dfrac{d}{dx}\:\left[\:\left(\:x^2\:+\:\sqrt{x^2\:+\:1}\:\right)^{\frac{1}{2}}\:\right]\:.\:\dfrac{d}{dx}\:\left(\:x^2\:+\:\sqrt{x^2\:+\:1}\:\right)}$

We know power rule of differentiation

$\displaystyle{\boxed{\pink{\sf\:\dfrac{d}{dx}\:(\:x^n\:)\:=\:n\:x^{n\:-\:1}\:}}}$

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

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

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:\dfrac{1}{2\:\sqrt{x^2\:+\:\sqrt{x^2\:+\:1}}}\:.\:\dfrac{d}{dx}\:\left(\:x^2\:+\:\sqrt{x^2\:+\:1}\:\right)}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:\dfrac{\dfrac{d}{dx}\:\left(\:x^2\:+\:\sqrt{x^2\:+\:1}\:\right)}{2\:\sqrt{x^2\:+\:\sqrt{x^2\:+\:1}}}}$

We know addition rule of differentiation,

$\displaystyle{\boxed{\green{\sf\:\dfrac{d}{dx}\:(\:u\:+\:v\:)\:=\:\dfrac{du}{dx}\:+\:\dfrac{dv}{dx}\:}}}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:\dfrac{\dfrac{d}{dx}\:\left(\:x^2\:\right)\:+\:\dfrac{d}{dx}\:\left(\:\sqrt{x^2\:+\:1}\:\right)}{2\:\sqrt{x^2\:+\:\sqrt{x^2\:+\:1}}}\:}$

By using power rule of differentiation,

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

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:\dfrac{2x\:+\:\left[\:\dfrac{1}{2}\:\left(\:x^2\:+\:1\:\right)^{\frac{1}{2}\:-\:1}\:.\:\dfrac{d}{dx}\:\left(\:x^2\:+\:1\:\right)\:\right]}{2\:\sqrt{x^2\:+\:\sqrt{x^2\:+\:1}}}\:}$

By using addition rule of differentiation,

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:\dfrac{2x\:+\:\left[\:\dfrac{1}{2}\:\left(\:x^2\:+\:1\:\right)^{-\:\frac{1}{2}}\:.\:\left(\:\dfrac{d}{dx}\:\left(\:x^2\:\right)\:+\:\dfrac{d}{dx}\:\left(\:1\:\right)\:\right)\:\right]}{2\:\sqrt{x^2\:+\:\sqrt{x^2\:+\:1}}}\:}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:\dfrac{2x\:+\:\left[\:\dfrac{1}{2}\:\dfrac{1}{\sqrt{x^2\:+\:1}}\:.\:\left(\:2x^{2\:-\:1}\:+\:0\:\right)\:\right]}{2\:\sqrt{x^2\:+\:\sqrt{x^2\:+\:1}}}\:}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:\dfrac{2x\:+\:\left(\:\dfrac{1}{\cancel{2}}\:\dfrac{1}{\sqrt{x^2\:+\:1}}\:.\:\cancel{2}x\:\right)}{2\:\sqrt{x^2\:+\:\sqrt{x^2\:+\:1}}}\:}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:\dfrac{dy}{dx}\:=\:\dfrac{2x\:+\:\dfrac{x}{\sqrt{x^2\:+\:1}}}{2\:\sqrt{x^2\:+\:\sqrt{x^2\:+\:1}}}}}}}$