Question

# Quality Question

@ Integrals

$\sf if \: \: y = (x+ \sqrt{1+x^{2}})^{n}$
Then Find the value of

$\sf{(1+x)^{2} \dfrac{d^{2}y}{dx^{2}}+ x\dfrac{dy}{dx}}$
Need Full Explanation ​

Answer 1

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♠ Given :-

$\mathtt{y = (x+ \sqrt{1+x^{2}})^{n}}$

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♠ To Find :-

$\mathtt{(1+x)^{2} \dfrac{d^{2}y}{dx^{2}}+ x\dfrac{dy}{dx}}$

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♠ Answer :-

$\huge\purple{ \bf n {}^{2} y}$

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♠ Step by step Solution :-

$\mathtt{ y = (x+ \sqrt{1+x^{2}})^{n}}$

Differentiating both sides w.r.t x

$\large\mathtt{\dfrac{dy}{dx} = n.(x+ \sqrt{1+x^{2}})^{n - 1} }  \\ \large\mathtt{ \times  \frac{d}{dx} (x+ \sqrt{1+x^{2}})} \\  \\ \large \mathtt{\dfrac{dy}{dx} = n.(x+ \sqrt{1+x^{2}})^{n - 1} }  \\ \large \mathtt{ \times  \frac{d}{dx} (x+ {(1+x^{2}  {)}^{1/2} }) {}^{} }  \\  \\ \large \mathtt{\dfrac{dy}{dx} = n.(x+ \sqrt{1+x^{2}})^{n - 1} }  \\ \large\mathtt{ \times (1  +  \frac{d}{dx}(1+x^{2}  {)}^{1/2})} \\  \\ \large\mathtt{\dfrac{dy}{dx} = n.(x+ \sqrt{1+x^{2}})^{n - 1} }  \\ \large \mathtt{ \times (1  +   \frac{1}{2} (1+x^{2}  {)}^{ - 1/2}\times 2x})$

$\large\mathtt{\dfrac{dy}{dx} = n.(x+ \sqrt{1+x^{2}})^{n - 1} } \\ \large \mathtt{ \times  \bigg(1  +  \frac{x}{ \sqrt{1 +  {x}^{2} } } } \bigg) \\  \\ \large \mathtt{\dfrac{dy}{dx} = n.(x+ \sqrt{1+x^{2}})^{n - 1} }  \\ \large\mathtt{ \times  \bigg(  \frac{  \sqrt{1 +  {x}^{2}  } + x }{ \sqrt{1 +  {x}^{2} } } } \bigg) \\  \\  \large=  \mathtt{ \dfrac{n.(x +  \sqrt{1 +  {x}^{2} } ) {}^{n} }{ \sqrt{ 1 +  {x}^{2} } } } \\  \\ \large \mathtt{  \frac{dy}{dx} =  \frac{ny}{ \sqrt{1 +  {x}^{2} } } }$

$\purple{: \longmapsto\boxed{\boxed{ \large\mathtt{ ( \sqrt{1 +  {x}^{2} } )  \dfrac{dy}{dx}  = ny}}}}--(1)$

$\large \bigstar \underline{ \pmb{ \mathfrak{ Differentiating \: both \:sides\bf \: w.r.t \: x}}}$

$\large\mathtt{({\sqrt{ 1 +  {x}^{2} }).  \dfrac{ {{d}^{2}y }}{ {{dx}^{2} }}}}     \\   \\  \large +  \mathtt{ \frac{dy}{dx}. \frac{d}{dx}( \sqrt{1 +  {x}^{2} } )   = n \frac{dy}{dx} }$

$\mathtt{ \sqrt{1 +  {x}^{2} }.  \dfrac{ {d}^{2} y}{dx {}^{2} }  \:  \:   }  \\  +  \mathtt{  \dfrac{dy}{dx}  . \frac{1}{ \not2}  \times  \frac{ \not2x}{ \sqrt{1 +  {x}^{2} } } \mathtt{ = n \frac{dy}{dx}} }$

$\large\bf \bigstar Multiplying \: Both  \:Sides \:by\:  \\ \large \bf \sqrt{1 +  {x}^{2} }$

We Get ,

$\large\mathtt{(1  + {x}^{2})  \dfrac{ {d}^{2} y}{dx {}^{2} } + x \dfrac{dy}{dx}  = n.(ny)} \\    \bf   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \{\because  {eq}^{n}  \:  \: (1) \}$

$\large\pink{ : \longmapsto\mathtt{(1  + {x}^{2})  \dfrac{ {d}^{2} y}{dx {}^{2} } + x \dfrac{dy}{dx}  = n {}^{2} y}}$

$\Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$

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Answer 2

Step-by-step explanation:

See Attachment And Get your Answer

Hope It helps u