Question

Differentiation of (√x+1/√x)²

Answer 1

GIVEN :–

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• A function $\rm{ \left( \sqrt{x} + \dfrac{1}{ \sqrt{x} } \right) }^{2}$

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TO FIND :–

• Differentiate form of given function = ?

SOLUTION :–

• Let the function –

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$\rm \implies y = { \left( \sqrt{x} + \dfrac{1}{ \sqrt{x} } \right) }^{2}$

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• Using identity –

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$\rm \implies { \left(a+b \right)}^{2} = {a}^{2} + {b}^{2} + 2ab$

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• So that –

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$\rm \implies y = { ( \sqrt{x})^{2} + \left( \dfrac{1}{ \sqrt{x} } \right) }^{2} + 2( \sqrt{x} ) \left( \dfrac{1}{ \sqrt{x} } \right)$

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$\rm \implies y = x+ \dfrac{1}{x} + 2$

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• We should write this as –

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$\rm \implies y = x+ {x}^{ - 1} + 2$

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• Using formulas –

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

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$\rm \to \dfrac{d(constant)}{dx} = 0$

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• Now differentiate with respect to 'x' –

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

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$\rm \implies \dfrac{dy}{dx} = 1+( - 1){x}^{( - 1 - 1)}+0$

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$\rm \implies \dfrac{dy}{dx} = 1 - {x}^{ -2}$

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$\rm \implies \large{ \boxed{ \rm \dfrac{dy}{dx} = 1 - \dfrac{1}{{x}^{2}}}}$

Answer 2

${ \huge {\underline {\underline {\rm{❥Given:}}}}}$

A function $\sf(x + \frac{1}{x} {)}^{2}$

❥To find:

Differentiation form of thegiven function;

Let the function be y

$\sf \: y = (x + \frac{1}{x} {)}^{2}$

${\underline{ \overline {\boxed{ \sf{Identity}}}} }\\ \\ \: \: \: \: \: \: {\boxed{\sf{( a+ b{)}^{2} = {a}^{2} + {b}^{2} + 2ab}}}$

So that;

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

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

❥Formula:

$\rightarrow \sf\frac{d(x {)}^{n} }{dx} = n {x}^{n} - 1\: \\ \\ \rightarrow \sf \frac{d(constant)}{dx} = 0$

$\sf \rightarrow \frac{dy}{dx} = \frac{d(x)}{d(x)} + \frac{d( {x}^{ - 1}) }{dx} + \frac{d(2)}{dx}$

$\implies \sf \frac{dx}{dx} = 1 + ( - 1) {x}^{ - 1 - 1} + 0$

${ \large { \color{red}{ \underline{ \overline \color{black} {\implies \sf \frac{dy}{dx} = 1 - \frac{1}{ {x}^{2} } }}}}} \:$

${ \large{ \underline {❥\sf{ \red{H} \color{magenta}{o} \color{lime}{p} \blue{e }\: \purple{i} \pink{t } \: \color{cyan}{h} \color{red}{e} \color{navy}{l} \color{blue}{p}{s} \: \color{pink}{y} \orange{o} \color{purple}{u}.....}}}}$