Question

(root x + 1/rootx)² differentiate with respect to x

Answer 1

Step-by-step explanation:

we get the abbreviation for the 2nd step by using the identity (a+b)^2 .This is the answer.Hope this helps.

Answer 2

$\displaystyle \sf{ \frac{d}{dx} \bigg[ {\bigg( \sqrt{x} + \frac{1}{ \sqrt{x} } \bigg)}^{2} \bigg] = 1 - \frac{1}{ {x}^{2} } }$

$\displaystyle \sf{ \bf{Given : } {\bigg( \sqrt{x} + \frac{1}{ \sqrt{x} } \bigg)}^{2} }$

To find : To differentiate

Solution :

Let y be given function

Then we have

$\displaystyle \sf{y = {\bigg( \sqrt{x} + \frac{1}{ \sqrt{x} } \bigg)}^{2} }$

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

$\displaystyle \sf{ \implies \: y = x + 2 + \frac{1}{x} }$

Differentiating both sides with respect to x we get

$\displaystyle \sf{ \: \frac{dy}{dx} = \frac{d}{dx} \bigg( x + 2 + \frac{1}{x} \bigg) }$

$\displaystyle \sf{ \implies \: \frac{dy}{dx} = \frac{d}{dx} \bigg( x \bigg) + \frac{d}{dx} \bigg( 2 \bigg) +\frac{d}{dx} \bigg( \frac{1}{x} \bigg) }$

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

$\displaystyle \sf{ \implies \: \frac{dy}{dx} = 1 + 0 - {x}^{ - 1 - 1} }$

$\displaystyle \sf{ \implies \: \frac{dy}{dx} = 1 - {x}^{ - 2} }$

$\displaystyle \sf{ \implies \: \frac{dy}{dx} = 1 - \frac{1}{ {x}^{2} } }$

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