Question

find dx/du when x=(u+1/u)²​

Answer 1

Given

$\tt \to \: x = \bigg(u + \dfrac{1}{u} \bigg)^{2}$

To Find dx/du

First of all Simplify the Equation

$\tt \to \: x = {u}^{2} + \dfrac{1}{ {u}^{2} } + u \times \dfrac{1}{u}$

$\tt \to \: x = {u}^{2} + \dfrac{1}{ {u}^{2} } + \cancel{u}\times \dfrac{1}{ \cancel{u}}$

$\tt \to \: x = {u}^{2} + \dfrac{1}{ {u}^{2} }$

Now dx/du is

$\tt \to \: \dfrac{d ( {u}^{2} + {u}^{ - 2} ) }{du}$

We know that

$\tt \to \: \dfrac{d(f(x) \pm \: g(x))}{dx} = \dfrac{d(f(x))}{dx} \pm \dfrac{d(g(x))}{dx}$

We Get

$\tt \to \: \dfrac{d( {u}^{2} )}{du} + \dfrac{d( {u}^{ - 2}) }{du}$

using Formula

$\to \tt \dfrac{d( {x}^{n} )}{dx} = n {x}^{n - 1}$

we get

$\tt \to \: 2u ^{2 - 1} + ( - 2 {u}^{ - 2 - 1} )$

$\tt \to 2u - 2 {u}^{ - 3}$

$\tt \to \: 2u - \dfrac{2}{ {u}^{3} }$

Answer

$\tt \to \: 2u - \dfrac{2}{ {u}^{3} }$